The First Derivative Test - Online Tutor, Practice Problems & Exam Prep

Critical Points

At this point in the course, we know how to find the local extrema of a function by looking at its graph. Like here for a function \( g(x) \), we can see that it has a local maximum at \( x = -4 \), a local minimum down here at \( x = -1 \), and finally, a local maximum when \( x = 4 \). But we won't always be given the graph of a function, and we'll still need to identify the local extrema when just given the function itself. But how exactly can we even start in doing that? Well, the first step in being able to find the local extrema of a function is being able to find what's called that function's critical points.

Now this is exactly what I'm going to walk you through here, what critical points are and how exactly to find them. So, let's go ahead and jump right in here. Now coming back over to our graph of \( g(x) \) here, if we look at all of these local extrema, they all have something in common if we think about their tangent lines. For this first local max and local min, if I draw a tangent line, I can see that it is a horizontal tangent line. And for this last local max over here, because it comes up to a point, I know that it has no tangent line at all and this is actually going to be true of all local extrema.

They all either have a horizontal tangent like those first two points or no tangent line at all. So why does this information actually matter? Well, if we know information about a function's tangent, then we also know information about that function's derivative because it's just the slope of the tangent. So if a function has a horizontal tangent line, I know that at that point, my derivative will be equal to 0, that zero slope of that horizontal tangent. And if my function has no tangent line at a point, I know that there, my derivative does not exist.

Now these points where a derivative equals 0 and where a derivative does not exist are what's called critical points. And remember earlier, I mentioned that finding critical points is going to be the first step in being able to find local extrema. So let's actually find the critical points of this function here, \( f(x) = x^3 - 12x + 5 \). And in order to find those critical points, that means we need to find where our derivative equals 0 and where our derivative does not exist.

So in order to do that, let's first just find what our derivative is. So \( f'(x) \) here using the power rule is going to be \( 3x^2 - 12 \). So now that I have that derivative, to find where it's equal to 0, I just need to set that \( 3x^2 - 12 = 0 \) and solve for x. Now adding 12 on both sides here, that will cancel out, leaving me with \( 3x^2 = 12 \). Then dividing both sides by 3 cancels that 3.

I'm left with \( x^2 = 4 \). And then finally, last step, taking the square root on both sides, I end up with \( x = \pm 2 \). So this gives me 2 critical points where \( x = 2 \) and where \( x = -2 \). But remember, this is just half of the story. We also need to find where a derivative does not exist.

And this is specifically where it does not exist for values of \( x \) that actually do exist in the domain of our original function. Now here, since we're working with a basic polynomial function, there is actually nowhere where this derivative does not exist. But it is something to keep in mind as you work with different types of functions like, say, rational functions or those with rational exponents. So here we already have all of our critical points. So we have 2 critical points where \( x = 2 \) and where \( x = -2 \).

So now that we have our critical points, there is something that I want to note here, and that's that not all critical points are local extrema. So what exactly does that mean? Well, it means that just because 2 and -2 are critical points, that doesn't automatically make them a local maximum or local minimum. Now, this can be a bit hard to understand, so let's come back over to our graph here. We know that these local extrema have either a horizontal tangent or no tangent at all.

But this point here at \( x = 2 \) actually also has a horizontal tangent and it's not a local max or min at all. So just because a function is a critical point, meaning that its derivative is equal to 0 or does not exist, that doesn't automatically make it a local maximum or local minimum. Finding critical points is just the first step in finding local extrema and also in finding a lot of other things out about a function. See, critical points are actually going to be used to find out a lot of different things, including local and global extrema, but also a lot of other stuff. Now, I'm going to walk you through absolutely everything that critical points are used for as we continue on in this course.

But for now, let's get some more practice with finding critical points. I'll see you in the next video.

Critical Points Example 1

In this problem, we're asked to find the critical points of the given function \( f(x) \). The function that we're given here is \( f(x) = \frac{x^2}{x-4} \). Now, remember that when we are asked to find critical points, that means that we want to find where our derivative is equal to 0 or where it does not exist. So, we need to start here by first finding the derivative of our function. Now, because we're working with a rational function—we have one function being divided by another—we need to use the quotient rule to find our derivative, \( f'(x) \), which, if you remember our memory tool for the quotient rule, we can think to ourselves, okay, "low d high, minus high d low over the square of what's below."

We're taking our bottom function, and we are squaring it. That's what goes on the bottom of our derivative. So, now that we have this derivative, we can do a bit of simplification here. We can go ahead and distribute the \( 2x \) into those parentheses. That gives me \( 2x^2 - 8x \) and then minus \( x^2 \) and then keeping that denominator the same, \((x - 4)^2\).

Now we can do further simplification here by going ahead and combining like terms. \( 2x^2 - x^2 \) is just going to give me \( x^2 \). So I have \( x^2 - 8x \) all over \((x - 4)^2\). And then finally, I can take one final step here, factor out an \( x \) in the top, leaving me with \( x(x - 8)/(x - 4)^2 \). It's always a good idea to go ahead and factor if you see something that's easily factorable because it will end up helping you in the long run.

Okay. We're not done yet. This is just our derivative. Right? And we want to find where this derivative is equal to 0 or where it does not exist.

Now, in finding where this derivative is equal to 0, we effectively just want to take those functions on the top, our numerator here, and set our numerator equal to 0. Since we have 2 factors that are multiplying each other, we have \( x \) and \( x - 8 \), we can set each of these individual factors equal to 0. This factorization makes it simpler. So, we can take the first factor, \( x = 0 \), and that is already a critical point. It's already solved for \( x \).

And then we can take \( x - 8 \), set that equal to 0. If I add 8 on both sides, that gives me my second critical point at \( x = 8 \). So now I have 2 critical points, one at \( x = 0 \) and one where \( x = 8 \). But remember that I also need to think about where my derivative does not exist. Because I'm dealing with a rational function, there is a point where this derivative does not exist.

Because if in my denominator I end up with a 0, meaning that \( x - 4 \) would be equal to 0, that is where my derivative wouldn't exist. Now if I add 4 to both sides here, that gives me \( x = 4 \). But I need to think about this a little more because the only critical points are going to be where it does not exist, but it is still defined for our function. Now our original function was \( x^2/(x - 4) \), meaning that my original domain restriction tells me that \( x \) cannot be equal to 4 because that would make my denominator 0. So even though my derivative does not exist for \( x = 4 \), this is not a critical point because my function also doesn't exist there.

So for this function, \( x^2/(x - 4) \), I have only 2 critical points: where \( x = 0 \) and where \( x = 8 \). Now feel free to let us know if you have any questions, and let's continue practicing.

Find the critical points of the given function.

$g(x)=\frac13x^3-\frac12x^2-12x$

• A. $x=0,x=12$
• B. $x=-4,x=3$
• C. $x=-6,x=2$
• D. $x=4,x=-3$

Find the critical points of the given function.

$f(t)=\frac{6t}{t^2+1}$

• A. $t=0,t=6$
• B. $t=1$
• C. $t=-1,t=1$
• D. No critical points

Find the critical points of the given function.

$f(x)=\sqrt{2-x^2}$

• A. $x=0$
• B. $x=0,x=\sqrt2,x=-\sqrt2$
• C. $x=0,x=\sqrt2$
• D. No critical points

Critical Points Example 2

Here we're asked to find the critical points of the given function. The function that we're given here is y=lnx-1. Now remember that in order to find critical points, we're looking for where our derivative, so in this case, y′, is either equal to 0 or does not exist. So the first thing we want to do here is find the derivative. So finding our derivative here, y′, we know that the derivative of the natural log is 1x over whatever the argument of that natural log is.

Now here, that is 1x-1. Using the chain rule here, I don't actually need to do anything because the derivative of that inside function is also 1. So my derivative is just 1x-1. So we want to find where this derivative is either equal to 0 or does not exist. Now, there are actually no values for which this is going to be equal to 0 because there's nowhere that our numerator is going to be equal to that.

So we don't need to even think about this being equal to 0. We instead need to focus on where it does not exist. And this is specifically going to be where our denominator is equal to 0 because we know that a denominator cannot be 0. So if I set that denominator x-1 equals to 0, that then gives me, adding 1 to both sides, x=1. So this may be a critical point.

But remember that whenever we're working with these, we need to make sure that they're inside of the domain of our original function. If I were to plug 1 into my original function, that would give me the natural log of 0, which is not something that I can do. So because this one is not within the domain of my original function, I just don't have any critical points. So for this function, the natural log of x-1, there are actually no critical points at all. Now feel free to let us know if you have any questions here.

Thanks for watching, and I'll see you in the next video.

Determining Where a Function is Increasing & Decreasing

If we look at the graph of a function from left to right, we know that a function is increasing if it goes up and decreasing if it goes down. But we won't always be given the graph of a function, and we still need to be able to determine where a function is increasing or decreasing just based on the function itself. And that's exactly what I'm going to show you how to do here, find where a function is increasing or decreasing by actually just using its derivative. I'm also going to walk you through a really common problem type that you're going to encounter step by step. So let's go ahead and get started.

Now looking at the graph of our function here, if I look where my function is increasing and draw some tangent lines, I can see that the slope of these tangent lines is positive. Then where my function is decreasing, the slope of these tangent lines becomes negative. Now by this point, we know all too well that the slope of a tangent line is just the derivative. So where my function is increasing, my derivative is positive. Positive.

We can see that it's above the x-axis here. And where my function is decreasing, my derivative is negative. So if I don't have any graphs at all, then I know that my function is increasing or decreasing based on the sign of its derivative. So in this problem, we're specifically asked to determine if a function is increasing or decreasing at these specific values of x, x equals 0 and x equals 5. And we can do that by finding the sign of the derivative at each of these points.

So let's start here by just finding the derivative of our function. Here we have the function f(x)=-x2+4x+5. So finding my derivative here using the power rule, f′(x)=-2x+4. Now in order to find the sign of my derivative for these individual points, I just need to plug them in. So f′(0), plugging that in, gives me negative 2 times 0+4, which gives me a positive 4.

Now I don't actually care what this value is. I just care about the sign of it. Now since this is a positive 4, whenever the derivative is positive, that tells me that my function is increasing there. So I know that my function is increasing at x equals 0 because at that point, the derivative is positive. So let's go ahead and plug in 5 to our derivative as well.

Now plugging 5 in f′(5) is going to be equal to -2 times 5+4. This gives me negative 6. Now again here, I just care about the sign of this value. This is negative and when your derivative is negative, you know that your function is decreasing. So we know that our function is decreasing at x equals 5.

So for specific values of x, determine where your function is increasing or decreasing based on the sign of the derivative at that point. If it's positive, then you know your function is increasing. If the sign of your derivative is negative, your function is decreasing. Now this was for specific values of x, but a more common thing that you'll actually be asked is to determine entire intervals for which a function is increasing or decreasing, and we're gonna walk through that together. Now we're gonna follow a super similar process to what we did up here just with a slightly different structure.

So let's take a look at our example here. We're going to be working with the same exact function, that's -x2+4x+5, And the very first thing that we want to do here when determining the intervals for which our function is increasing or decreasing is actually finding the critical points of that function. Now remember that the critical points are where the derivative of your function is either equal to 0 or does not exist. Now we actually already found our_derivative up above. We saw that f′(x) was equal to -2x+4.

Now since this is a basic polynomial function, I know that there's nowhere where this derivative won't exist. So I can find critical points by just setting this equal to 0. Now solving for x here subtracting 4 on both sides, I get -2x=-4. Then dividing both sides by negative 2, I get x equals 2. So I only have a one critical point at x equals negative 2.

So I'm done with step 1, and we can move on to step 2. This is where we're actually going to determine what these intervals that we're talking about are, and we're going to determine those intervals using what's called a sign chart. Now a sign chart is literally just a number line, and we're going to create this number line, create these intervals on our number line using our critical points that we just found in step 1. Now we only have one critical point here, x equals 2, and I'm gonna go ahead and mark that on my sign chart 2, and that then splits my sign chart into 2 intervals, from negative infinity until I reach that 2 and then from 2 on to infinity. Now note that these are open intervals.

Our critical point is not going to be included in either interval. Once we have these intervals, we want to determine if our function is increasing or decreasing over each of them. And we can do that by plugging a value from each interval into our_derivative. Now this is the exact same thing that we did up above, plug specific values into our_derivative and take a look at the sign. So we want to do the same thing here.

The only difference is we now have to choose what those values are. Now when you're choosing these sort of test values, you really just want to choose them so that it makes it easy to actually plug back into your_derivative. It doesn't actually matter what they are. They can be any value within this interval. Now from negative infinity to 2, I'm gonna go ahead and choose 0.

That's usually a safe bet. And then from 2 to infinity, I'm gonna go ahead and choose 5. Remember, I can choose any value here. Now from here, we just want to plug these values into our_derivative. So I want to find f′(0) and f′(5).

But I actually already plugged these values into my_derivative up above. Remember that f′(0) was positive 4 and f prime of 5 was negative 6. So since I already have that here, that's positive 4 and negative 6. In the same way that up above, we only cared about the sign of these values, the same thing is true here. See, if our value is positive, like it is right here when x is equal to 0, we then know that our function is increasing on that entire interval.

So based on that value being positive, based on our_derivative being positive here, we know that our function is increasing on this entire interval from negative infinity to 2. Now you may be wondering why this is for the whole interval and not just for that one single value that we plugged in. And that's because when we created our sine intervals using our critical points, we already took care of that. And this means that every single value, no matter what value I choose, if I plug it into my_derivative, it's going to give me a positive value because my function is increasing on that entire interval. So let's take a look at f′(5).

We saw that here our_derivative was negative. And when our_derivative is negative, that tells us then that our function is decreasing again on that entire interval. So we know that our function is decreasing from 2 to infinity, and that's all there is to it. In order to determine where a function is increasing or decreasing, the first thing you want to do is find your critical points. Then you want to make a sign chart and find those intervals based on those critical points you found.

From there, you can choose a test value inside each interval. Plug that into your_derivative. If your_derivative there is positive, your function is increasing. If it's negative, your function is decreasing. So now that we know how to determine the intervals for which a function is increasing or decreasing, let's continue getting practice with it.

I'll see you in the next video.

Identify the open intervals on which the function is increasing or decreasing.

$f(x)=3x^4+8x^3-18x^2+7$

• A. Increasing on $\left(-\infty,-3\right)\&\left(0,1\right)$, Decreasing on $\left(-3,0\right)\&\left(1,\infty\right)$
• B. Increasing on $\left(-3,0\right)\&\left(1,\infty\right)$, Decreasing on $\left(-\infty,-3\right)\&\left(0,1\right)$
• C. Increasing on $\left(-\infty,0\right)$, Decreasing on $\left(0,\infty\right)$
• D. Increasing on $\left(0,\infty\right)$, Decreasing on $\left(-\infty,0\right)$

Identify the open intervals on which the function is increasing or decreasing.

$f(x)=x^{2/3}(4-x)$

• A. Decreasing on $\left(-\infty,\infty\right)$
• B. Increasing on $\left(-\infty,\infty\right)$
• C. Increasing on $\left(-\infty,0\right)\&\left(\frac85,\infty\right)$, Decreasing on $\left(0,\frac85\right)$
• D. Increasing on $\left(0,\frac85\right)$, Decreasing on $\left(-\infty,0\right)\&\left(\frac85,\infty\right)$

Identify the open intervals on which the function is increasing or decreasing.

$f(x)=xe^{-2x}$

• A. Increasing on $\left(-\infty,\frac12\right)$, Decreasing on $\left(\frac12,\infty\right)$
• B. Increasing on $\left(\frac12,\infty\right)$$\left(-\infty,\frac12\right)$, Decreasing on $\left(-\infty,\frac12\right)$
• C. Increasing on $\left(-\infty,\infty\right)$
• D. Decreasing on $\left(-\infty,\infty\right)$

Determining Where a Function is Increasing & Decreasing Example 3

In this problem, we're asked to identify the open intervals where our function \( f(x) \) is increasing or decreasing based on the graph here of \( f'(x) \). It's really important here to notice that we're dealing with the graph of a derivative rather than the graph of our function itself. So we're not looking for where this graph is going up or down. We're looking for where it's positive or negative because I know that where \( f'(x) \) is positive, my original function \( f(x) \) will be increasing. And where \( f'(x) \) is negative, my original function \( f \) will then be decreasing.

So that's all I'm looking for here, where this graph is above the x-axis versus below the x-axis. That will tell me where my original function \( f(x) \) is increasing or decreasing. So let's take a look at this graph from left to right. Now here from negative infinity until I reach that negative five, I can see that my graph is below the x-axis. It is negative.

So that tells me that my original function \( f \) is then going to be decreasing from negative infinity until we reach that negative 5. At negative 5, my graph then becomes positive. So my derivative becomes positive here. Then it's above the x-axis all the way until we reach positive 3.

So that tells me that my function \( f \) is increasing from negative 5 all the way to positive 3. Then at 3, our graph dips back below the x-axis. It is negative again until we reach positive 5. So here my function is decreasing again from 3 to 5. Then finally at 5, we continue on above the x-axis.

So that tells me that because my derivative there is positive, my function \( f \) will be increasing again from 5 on to infinity. So whenever you're given a graph and asked to determine something like where your function is increasing or decreasing, be sure to pay attention to what type of graph you're dealing with. Here we are given the graph of a derivative, so we only want to look at where it's positive or negative to tell us where our original function \( f \) is increasing or decreasing. Let us know if you have any questions here, and I'll see you in the next one.

The First Derivative Test: Finding Local Extrema

Hey everyone! Recall that in the previous video, we learned that the sign of the derivative tells us whether a function is increasing or decreasing, but it can actually tell us even more than that if we pay attention to where that sign actually changes. Where the sign of our derivative changes tells us where the local extrema of our function occur. That's our local maximum and minimum values that we know often look like these sort of peaks and valleys on a graph here. Now, in order to find our local maxima and minima using the sign changes of our first derivative, we are going to use what is called the first derivative test.

Now here, I'm going to walk you through exactly what the first derivative test is and how to use it to find the local extrema of a function step by step. So let's go ahead and get started. Just as we did when finding where a function was increasing or decreasing, we want to start here from our critical points. Remember that our critical points are where that derivative is either equal to 0, like it is for these three points on our graph here, or where that derivative does not exist. In order to use our first derivative test, we want to look at what's happening with the sign of our derivative on either side of our critical points.

Now for this first critical point here, my function starts off increasing. That means it has a positive derivative. Then it begins decreasing on the other side of that critical point, meaning that derivative is negative. Where this derivative sign changes from positive to negative, that tells me that my function has a local maximum located at that critical point, which we can see is exactly what happens at this critical point on our graph here. Now, looking at that second critical point, that function is decreasing and has that negative derivative, then it begins increasing again, that derivative becomes positive.

Where that derivative sign changes from negative to positive, that tells us then that our function has a local minimum located at that critical point, which again, we can see is exactly what happens for this critical point here. Now for that very last critical point, we can see that our function is increasing, has a positive derivative, and it continues increasing, continuing on with that positive derivative. So the sign did not change at all. And whenever the sign of our derivative does not change, that just means that our function will have no local extrema there, which we can see again is exactly what happened on our graph view. So now that we've walked through what the first derivative test actually tells us, let's use it to find the local extrema of a function.

Here, we're given the function f(x)=x3−3x2+4. Now these steps are going to look really familiar because they're almost identical to how we found where a function was increasing or decreasing. Now we're just going to apply that first derivative test once we get to that 3rd step. Our first step here is, of course, to find our critical points. That's where our first derivative is either equal to 0 or does not exist.

We actually already have our first derivative worked out for us here. It is 3x2−6x or factored to 3x(x−2). There's nowhere this derivative won't exist, so we just need to set it equal to 0. That means that we can set each of these individual factors equal to 0 since they're multiplying each other. So I have 3x=0 and x−2=0.

Solving for x here, that 3 will cancel, and I'm left with that first critical point at x=0. Then for my second critical point, if I add 2 to both sides, that will cancel out and leave me with x=2. So with those critical points found, I now want to use them to make my sign chart intervals. I'm going to go ahead and mark them on my number line down here as I've done before, 0 and 2, and use that to split into intervals. So I have 3 intervals here from −∞;0, from 0 to 2, and finally from 2;∞.

So with steps 1 and 2 done, we move on to step 3, which is where we want to go ahead and plug a test value from each interval into our first derivative here. The same way we've chosen a test value 聴before, remember we want to choose values that are going to make it easy to plug back into our derivative. So in that first interval, I'm going to go ahead and choose -1 and then from 0 to 2, I'm going to choose positive 1 and then from 2 to infinity, I'm just going to choose 3. Remember, you can choose whatever numbers make it easiest for you. Now from here, I want to plug these into that first derivative.

So I want to find f′(−1), f′(1), and f′(3). Now you can go ahead and pause here and work this out on your own before checking back in with me. When you plug negative one into that first derivative, that 3x(x−2), you should get a positive 9. Then if you plug positive 1, you should startX receiving negative 3. Then finally, if you plug 3 in you should again end up with a positive 9.

Now from here is where we actually want to use our first derivative test. This down here in my steps box is just a restatement of exactly what we learned up above. Just as we were before, again, here we only care about the sign of that derivative here. So when we calculated those values, really all we need to know is the sign. For that negative one, we got that positive 9.

This is a positive value, so that means my first derivative is positive over this whole_interval_So over that second interval, we got a negative value. And then finally, over that last_interval_, we got a positive value. Now applying our first derivative test here, we wanna look at where the sign of our derivative changes. From our first interval to our second interval, we can see that that derivative_changes_from being positive, our function increasing, to being negative, so our function decreasing.

Now, when this happens based on our first derivative test, that tells us that our function has a local maximum at that point. So based on that, I know that my function then has a_local_maximum_here_at that_critical_point 0, so_x_equals_0_. Then looking at my second interval to my 3rd_interval_, this change_here_, I can see that my first_derivative_goes from being negative to then being positive. My function goes from decreasing to increasing. And when this_Happens based on_our first derivative test, I know that my function then has a local_minimum there.

So I know that my function has a local minimum at x=2. So having located the local extrema of this function, this function f(x) has a local maximum at x=0 and a local minimum at x=2, just based_on the_sign_changes_of_our_derivatives. Now here, we were just asked to locate our local extrema. But if you're actually asked to find the value of these maximum and minimum points, you wanna go ahead and plug your critical points back into your original function f(x) = x3−3x2+4 in order to get those_maximum_and_minimum_values.

But here, we're_already_done. We found our local max and min_based_on_our first_derivative_test. Feel free to let me know if you have any questions here, and let's continue practicing in the next video.

The First Derivative Test: Finding Local Extrema Example 4

In this problem, we're asked to identify the local minimum and maximum values of the given function, if there are any. The function that we're given here is f(x)=x+7)3. Remember that in order to find our local extrema, we can use the first derivative test where we want to find our critical points and then test if our derivative changes at that critical point. Then we can see, based on that change, whether that point is a local max, a local min, or neither. So let's go ahead and get started here.

Our first step is to find our critical points. This is where our derivative is either equal to 0 or does not exist. So we want to start by just finding our derivative, f′(x). Using the power rule, that gives me 3×(x+7)2. And then I want to multiply by the derivative of what's inside, which just happens to be 1, so that's not going to change anything.

Now there's nowhere that this derivative won't exist, so I can go ahead and just set it equal to 0 to find my critical points. Doing so, I have x+7=0. Subtracting 7 on both sides gives me my single critical point at x=−7. With this critical point, I can make my sign chart intervals, marking negative 7, and then splitting this into my intervals.

This goes from negative infinity until we reach that critical point at negative 7 and then from negative 7 on to positive infinity. Now that I have those intervals, I want to choose a test value from each interval to plug into the first derivative and check the sign changes, if there are any, at that point. Let's choose a test value in each of our intervals here. From negative infinity to negative 7, I'll choose negative 8. From negative 7 to infinity, I'll choose 0, to plug these values in.

Finding f′(−8) and also f′(0), plugging in negative 8 into our derivative, we get 3×(−1)2=3, which is just a positive 3. This indicates that the derivative is positive. Plugging in zero gives us 3×49=147, which is also positive.

The derivative is positive in both intervals, indicating that it starts out increasing and at that critical point, it continues increasing. The sign of our derivative doesn't change at our critical point. Our first derivative test tells us that if the sign does not change, then we have no local extrema for this function. We have no local extrema for this function at all, and that is our final answer here. This makes sense given the nature of a cubic function, where we typically see continued increase throughout. There is no local minimum or maximum at this critical point. Feel free to let us know if you have any questions, and I'll see you in the next video.

The First Derivative Test: Finding Local Extrema Example 5

If you haven't already tried to solve this problem on your own, feel free to pause here and do that before checking back in with me. Now, here we're asked to find the local minimum and maximum values of the given function, if there are any. Now, the function that we're given here is \( h(t) = \frac{t^3}{2t + 1} \). Remember that we can find our local minimums and maximums by using our first derivative test, checking where there are sign changes of that first derivative. The first thing we want to do here is find our critical points so that we can then test the sign of our first derivative on either side of those critical points.

So the first thing we want to do here is find where our derivative is either equal to 0 or does not exist, which means we want to find our derivative here, in this case, \( h'(t) \) since our original function is \( h(t) \). Given that we have one function dividing another function, we'll use our quotient rule. Remember our memory tool for the quotient rule, "low d-high minus high d-low over the square of what's below." We'll take the lower function \( 2t + 1 \), multiply it by the derivative of the upper function \( t^3 \), which using the power rule is \( 3t^2 \), minus \( t^3 \) times the derivative of \( 2t + 1 \), which is just 2, all over the square of what's below.

That's \( (2t + 1)^2 \). We can simplify a bit by distributing that \( 3t^2 \), giving us \( 6t^3 + 3t^2 - 2t^3 \), with the denominator remaining the same, \( (2t + 1)^2 \). Combining like terms produces \( 4t^3 + 3t^2 \) all over \( (2t + 1)^2 \).

If I factor \( t^2 \) from my numerator, that gives \( t^2(4t + 3) \) all over \( (2t + 1)^2 \). Here is my derivative, but this isn't our ultimate goal. We want to find where this derivative is equal to 0 or does not exist.

To find where it's equal to 0, we set the numerator equal to 0, effectively setting each factor equal to 0. So, \( t^2 = 0 \) gives \( t = 0 \) as the first critical point. For \( 4t + 3 = 0 \), subtracting 3 and dividing by 4 gives \( t = -\frac{3}{4} \) for the second critical point.

But remember, we also need to find where it does not exist, which is where the denominator equals 0. Setting \( 2t + 1 = 0 \) gives us another potential critical point at \( t = -\frac{1}{2} \). Since the denominator of our original function, \( h(t) \), is \( 2t + 1 \), our function does not exist at this point, and thus it cannot be a local extremum value. So we ignore it, focusing only on the two critical points.

We proceed to our second step by making our sign chart intervals based on our critical points, marking \( t = 0 \) and \( t = -\frac{3}{4} \). We then decide the sign of \( h'(t) \) by testing values in the intervals—choosing simple representatives like \( -1, -\frac{1}{4}, \) and \( 1 \). After plugging these values into \( h'(t) \) to determine the concavity, we find a switch from negative to positive derivative at \( t = -\frac{3}{4} \), indicating a local minimum.

Finally, we plug \( t = -\frac{3}{4} \) back into our original function \( h(t) \) to find the actual minimum value \( h(-\frac{3}{4}) = \frac{27}{32} \). Thus, the function has a local minimum value of \( \frac{27}{32} \) at \( t = -\frac{3}{4} \).

Identify the local minimum and maximum values of the given function, if any.

$f(t)=t^2\ln t$, $t>0$

• A. Local maximum of $1$ at $x=1$, Local minimum of $-0.19$ at $x=\frac12$
• B. Local maximum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local minima
• C. Local minimum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local maxima
• D. No local extrema

The First Derivative Test: Finding Local Extrema Example 6

In this problem, we're asked to identify the global and local minimum and maximum values of the given function, if there are any. Now here, we're given the function f(x) = x² + 4x + 17. Now because we are not given a closed interval here, you may be a bit confused as to how we're going to identify any global extrema. But this is where our knowledge of functions kicks in because we know that the function that we're working with is a parabola. So let's go ahead and follow the steps to find our local extrema and then circle back to that global question later.

So looking at our function here, f(x) = x² + 4x + 17, remember, we can find our local extrema by using the first derivative test, where we're going to test the sign of our first derivative on either side of our critical points. So we want to start here by finding those critical points. Now this is where our derivative is either equal to 0 or does not exist. So finding our derivative here using the power rule, this is going to be 2x + 4. Now there's nowhere this derivative doesn't exist, so I can just go ahead and set it equal to 0.

Subtracting 4 on both sides, I get 2x = -4. Then dividing both sides by 2, I get my critical point at x = -2. So I only have one critical point here. So when I set up my sign chart, I just have to mark that -2. This splits my function into its 2 intervals from negative infinity until we reach that critical point at -2, and then from -2 on to infinity.

So with that second step done, since we only have 2 intervals, we only have to choose 2 test values in this 3rd step, step 3, where we're plugging our values from each interval into that derivative. So on this interval from negative infinity to -2, I'm gonna go ahead and just choose -3. And then from -2 to infinity, I'm going to choose 0. Now we want to plug these values into that first derivative, so we're finding f prime of -3 and f prime of 0. So with those test values in mind, we have 2(-3) + 4 and then 2(0) + 4.

Now remember, we don't care what the values actually work out to be. We just care whether they're positive or negative. So looking at this, I can already see that this value is going to be negative, and I can see that this value will be positive. So looking at my 2 intervals here, on that first interval, I can see that my first derivative is negative. Then it becomes positive as we get to that other interval.

So this is where our first derivative test comes in, where we want to actually pay attention to what exactly is happening with the sign of our derivative as we cross from one side of our critical point to the other. Now on this side from negative infinity to -2, where that derivative is negative, we know that that means our function is decreasing. Then on the other side of that critical point, our derivative becomes positive, meaning that our function is now increasing. Because of this, at this critical point right here, we have a local minimum. So we know that we have a local minimum at x = -2.

So we actually want to find what this value is here by plugging it back into our original function. So with that local minimum identified, we want to go ahead and plug it back in to our original function here. So we want to find f(-2). Now when we do that, we can look back at our original function here, x² + 4x + 17. So this is going to be (-2)² + 4(-2) + 17.

This ends up working out to be 13. So that tells us that our function has a local minimum value of 13 at x = -2, that critical point. So this is our answer here for our local extrema. But remember that our original question asked us to find both global and local extrema. So we have our local extrema here, but what about our global extrema?

Remember that a point can be both a local and a global extrema value. Now looking at what is actually happening with our function here, we have the function x² + 4x + 17. This is a parabola. And further, because this value, this x², is positive, I know that I have an upward-facing parabola. So that local minimum value that we just found is the very bottom of that parabola. Because either side of that parabola goes up onto infinity, we know that this point is not only a local minimum but also a global minimum.

So this is both a global and local minimum of 13 at x = -2. When you're doing all of this math, don't forget about what you already know about your function. I hope this is helpful, and I'll see you in the next one.