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

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)=−2⋅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)=−2⋅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 there your function is decreasing. So we know that our function is decreasing at x equals 5.

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 negative x squared plus 4x plus 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 fprime(x)=−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 negative 2x equals negative 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 a 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 sorts 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)=4 was positive and f′(5)=−6 was negative. 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 intervals on which the function is increasing or decreasing.

$f(x)=\sin^2⁡x$ on $[0,\pi]$

• A. Increasing on $[0,π]$
• B. Decreasing on $[0,π]$
• C. Increasing on $[0,\frac{\pi}{2})$, Decreasing on $(\frac{\pi}{2},\pi]$
• D. Increasing on $(\frac{\pi}{2},\pi]$, Decreasing on $[0,\frac{\pi}{2})$

Determining Where a Function is Increasing & Decreasing Example 1

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 very important to note that we're dealing with the graph of a derivative rather than the graph of our function itself. We're not looking for where this graph is going up or down. We're looking for where it is 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, I'm just looking for 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. Let's take a look at this graph from left to right. From negative infinity until I reach negative five, I can see that my graph is below the x-axis. It is negative.

This indicates that my original function \( f \) is decreasing from negative infinity until we reach negative five. At negative five, my graph then becomes positive. The derivative becomes positive here. It remains above the x-axis all the way until we reach positive three.

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

This indicates that because my derivative is positive, my function \( f \) will be increasing again from five onwards to infinity. 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're given the graph of a derivative, so we only want to look at whether it's positive or negative to determine 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.

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. Now 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. Looking at that second critical point, that function is decreasing and has that negative derivative, then it begins increasing again, that derivative becomes positive.

Now 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. 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 here. 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. These steps are going to look really familiar because they are 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 third 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. Now that means that we can just 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 equals 0. Then for my second critical point, if I add 2 to both sides, that will cancel out and leave me with x equals 2. So with those critical points found, I now want to use them to make my sign chart intervals. So, 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 negative infinity to 0, from 0 to 2, and finally from 2 to infinity.

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 that we've chosen a test value 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 negative one 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 prime of negative 1, f prime of positive 1, and f prime of 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 get 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. Then 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 want to 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 local maximum at 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 my 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 equals 2. So having located the local extrema of this function, this function f(x) has a local maximum at x equals 0 and a local minimum at x equals 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 want to 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 minimum 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 2

In this problem, we're asked to identify the local minimum and maximum values of the given function, if there are any. Now, the function that we're given here is f(x)=(x+7))3. Now remember, 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 here by just finding our derivative, f'(x). Now 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. Now in doing so, this 3 will end up canceling and I just have x+7=0. That squared will also cancel, so I really just have x+7=0. And then if I subtract 7 on both sides, that gives me my single critical point at x=-7. So with this critical point, I can go ahead and make my sign chart intervals, marking that negative 7 down here, and then going ahead and splitting this into my intervals.

This goes from negative infinity until we reach that critical point at negative 7 and then from a negative 7 on to positive infinity. Now that I have those intervals, I want to go ahead and choose a test value from each interval to plug into that first derivative. And then I want to check the sign changes if it changes at all at that point. So let's go ahead and choose a test value in each of our intervals here. Now with this interval from negative infinity to negative 7, I'm gonna go ahead and choose a negative 8.

Remember, we're just choosing values that are going to make it easiest for us to plug in. If you feel like there's an easier value, you could have chosen something different and it's totally fine as long as it's within that interval. Then from my interval from negative 7 to infinity, I'm gonna go ahead and choose 0 to plug these values in. So I want to find f'(-8) and also f'(0). Now plugging in negative 8, remember we're plugging into our derivative here. Be sure not to get all of the functions we're working with mixed up here. We're plugging this into our derivative f', which here is 3×(x+7)2. So I want to go ahead and plug that in.

That's 3×(-8+7))2. That ends up giving me a 3×1, which is just a positive 3 because that becomes 3×1. So this gives me a positive 3. So here, my derivative is positive.

And then for 0, if we plug that in, that's going to be a 3×(0+7))2, which ends up being 3×49, which gives me a positive 147. So, again, this value is positive. So my derivative is positive on this interval and positive on this interval. That means that it starts out increasing, and at that critical point, it continues increasing. So our sign of our derivative doesn't change at this critical point at all.

Remember that our first derivative test tells us that if the sign does not change of that first derivative then we have no local extrema there. So we have no local extrema for this function at all, and that's our final answer here. Now this makes sense if we think about what type of function we're dealing with. We have a cubic function which often looks something like this. So this critical point is likely a point that looks like this where our function is increasing and then continues increasing on the other side.

No local minimum or maximum there. 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 3

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 of t equals t32t+1. Now remember that we can find our local minima and maxima by using our first derivative test, checking where there are sign changes of that first derivative. Now 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, so that means we want to find our derivative here, in this case, h prime of t since our original function is h of t. Now since I have one function dividing another function that means I want to go ahead and use my quotient rule. Remember our quotient rule, low d high minus high d low over the square of what's below. That's what we're going to apply here. So we're going to take that low function 2t+1, multiply it by the derivative of that t³, which using the power rule is going to be 3t², minus t³ times the derivative of that 2t+1, which is just 2 over the square of what's below.

That's 2t+1 squared. Now we can do a little bit of simplification here. If I distribute that 3t², that'll give me 6t³ plus 3t² minus 2t³, keeping that denominator the same here, which is 2t+12. Now we can go ahead and actually combine like terms, that negative 2t³ and that 6t³. That ends up giving me here 4t³ plus 3t² all over 2t+1 squared.

Now I can do one final bit of simplification here. If I go ahead and factor a t² out of my numerator here, that will give me t² times 4t+3 all over 2t+1 squared. So here is my derivative here, but this is not our ultimate goal. Right? We want to find where this derivative is equal to 0 or does not exist.

Now to find where it's equal to 0, I can go ahead and take my numerator here and set that equal to 0, which means that effectively, I can set each of these individual factors equal to 0. So that means that I'm looking for where t² equals 0 and also where 4t+3 equals 0. Now in doing that, this gives me t equals 0 for that first critical point. And then here, if I subtract 3 from both sides and then divide by 4 to cancel that 4 out. That ends up giving me t equals a negative 3 fourths for that second critical point here.

Now this is just where that derivative is equal to 0. But, remember, I also want to find where it does not exist. And since here I am working with a rational function, I can find where this does not exist by taking my denominator and setting that equal to 0. So taking that denominator, 2t+1 equals 0.

If I subtract 1 from both sides, that will cancel out, then I'm going to divide both sides by 2. So this gives me a third potential critical point at t equals negative one-half. Now I say potential critical point because if we think about what's happening with our actual function here, we can see that the denominator of our original function, h of t, is 2t+1. That's exactly what we just set equal to 0 here because the denominator of our derivative was just that 2t+1 squared. So because of that, that means that our actual function does not exist at this point.

Because of that, I can't have a local max or a local min there. Right? That wouldn't make any sense. So this point cannot be a local extremum. So I don't have to worry about it at all, and I can just work with these 2 critical points here.

So we can proceed to our second step here and go ahead and make our sine chart intervals based on our critical points. So making my sign chart here, I want to mark my 2 critical points that I am using here. That's t equals 0 and t equals a negative 3 fourths. My function actually exists at these points. So I have 0, I have negative 3 fourths.

I can split my sine chart into intervals from negative infinity to negative 3 fourths, from negative 3 fourths to 0, and from 0 on to infinity. Now that we have our sine chart, we want to choose a test value from each interval to plug into that derivative here, h prime. Now at this point, we want to choose those test values. Remember, we're choosing them such that it makes it easy for us to plug back into our derivative. Since our derivative is a little bit messy, these values aren't going to be super easy regardless of which values we choose, but we can still choose ones that make it semi easier than other values.

So from this

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

$f\left(\theta\right)=\sin\theta+\cos^2\theta$ on $[0,π]$

• A. Local maxima of $\frac54$ at $x=\frac{\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}$
• B. Local maxima of $\frac54$ at $x=\frac{\pi}{6}\&\frac{5\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}$
• C. Local maxima of $\frac54$ at $x=\frac{\pi}{6}$, Local minimum of $1$ at $x=\frac{\pi}{2}\&\frac{3\pi}{2}$
• D. No local extrema

The First Derivative Test: Finding Local Extrema Example 4

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 \( f(x) = x^2 + 4x + 17 \). Since we are not given a closed interval here, you may be a bit confused about 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 we are 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.

Looking at our function here, \( f(x) = x^2 + 4x + 17 \), 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. We want to start by finding those critical points. Now, this is where our derivative is either equal to 0 or does not exist. Finding our derivative here using the power rule, this is going to be \( 2x + 4 \). There is nowhere this derivative doesn't exist, so I can 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 \( x = -2 \). I only have one critical point here. When I set up my sign chart, I just have to mark that \( x = -2 \). This splits my function into two intervals: from negative infinity to negative 2, and then from negative 2 to infinity.

With that second step done, since we only have two intervals, we only have to choose two test values in this third step, where we're plugging our values from each interval into that derivative. For the interval from negative infinity to negative 2, I'm going to choose negative 3. And from negative 2 to infinity, I'm going to choose 0. We want to plug these values into that first derivative, finding \( f'( -3) \) and \( f'(0) \). 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. Looking at this, I can already see that this value is going to be negative, and this value will be positive. So looking at my two intervals here, on the first interval, I can see that my first derivative is negative. Then it becomes positive as we reach the other interval.

At this critical point, we have a local minimum since our function is decreasing before the point and increasing after it. We want to find \( f(-2) \). Plugging it back into our original function, \( (-2)^2 + 4(-2) + 17 \), we find 13. So our function has a local minimum value of 13 at \( x = -2 \).

But remember that our original question asked us to find both global and local extrema. Since \( x^2 \) is positive, we have an upward-facing parabola, and the local minimum value we just found is the very bottom of that parabola. Because either side of that parabola goes up to infinity, we know that this point is not only a local minimum but also a global minimum.

Thus, the function \( f(x) = x^2 + 4x + 17 \) has 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.