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

The Second Derivative Test: Finding Local Extrema

Earlier in this chapter, we learned how to find the local extrema of a function using the first derivative test. But it turns out that there's another way to find local extrema, using what's called the second derivative test, and specifically using the sign of the second derivative. Now here, I'm going to show you exactly what the second derivative test is and how to use it to find the local extrema of a function using a step-by-step process. So let's go ahead and jump right in here. Remember that the sign of our second derivative tells us whether a function is concave up or concave down.

And using concavity, we can determine whether a critical point is a local maximum or local minimum. But how exactly does this work? Let's take a look at our second derivative test here. If we have some point c and our first derivative of c is 0, meaning it is a critical point like we see on these two points on our graph here, if we then take a look at the second derivative, specifically the sign of our second derivative, if our second derivative is negative, that tells us that our function then has a local maximum at that critical point c. Now that might seem a little bit confusing.

Something being negative means that we have a maximum. But remember that when the sign of our second derivative is negative, that means that our function is concave down there. So looking at our function, it makes sense that where this function is concave down and has that critical point, this point would then be a local maximum, which is exactly what we see on our graph here. So then if our second derivative is instead positive, that indicates to us that our function has a local minimum at that point c because when our second derivative is positive, our function is concave up. So that critical point down there at the bottom is a local minimum.

Now, if our second derivative is not positive or negative at all, it's actually just 0, then that doesn't give us any information, and we instead need to revert back to using the first derivative test. But now that we have seen what the second derivative test is, let's actually use it to find the local extrema of a function. Now here, we're given the function f(x) = x³ - 3x² + 4. We want to locate the local extrema of this function using the second derivative test. Now the first step in doing this is actually finding our first derivative.

So it's important that we stay organized here and know exactly what function, what derivative, what second derivative we are working with here. Here, we want to find where that first derivative is equal to 0. So we want to find that first derivative f'(x), which you can do here using the power rule that gives us 3x² - 6x. If I go ahead and factor that here, if I factor 3x out of that, that's 3x(x - 2). Now I want to find where this is equal to 0, which I can do by setting each of these individual factors equal to 0.

So 3x = 0 and x - 2 = 0. Solving for x here, that 3 will cancel and leaves me with this point, x = 0. And then if I add 2 to both sides here, this gives me another point where that first derivative is equal to 0 at x = 2. Now one thing that you may notice here is we're used to finding our critical points where our first derivative is equal to 0 and where it does not exist. Now when working specifically with the second derivative test, we only want to find where that first derivative is equal to 0.

Because if it does not exist, we want to go ahead and revert back to using that first derivative test. But here, there are no points where that first derivative doesn't exist. So we are done with step 1, and we can proceed on to step 2. Now in step 2, we're going to use those values that we just found in step 1 and plug them into our second derivative, then taking a look at the sign of our second derivative. This is exactly where we'll be applying that second derivative test.

So first, just finding what that second derivative is here, f''(x). Again, using the power rule here, this gives me 6x - 6 or factored 6(x - 1). Now we want to plug these values in, x = 0 and x = 2, to this second derivative. So that means that I want to find f''(0) and f''(2), plugging those values in. Now f''(0) here, if I plug that in, that's going to be 6 * 0 - 1.

This gives us a negative 6. Now as we've seen before, we don't actually care what this value is. We just care about its sign. Now here, this is a negative value. Because here, our second derivative is negative, that tells us that this point is a local maximum.

This is exactly what we learned with our second derivative test above. So that tells us that this point is a local maximum located at x = 0. Remember, this negative six value doesn't tell us anything. We were only looking at the sign. Our local max is located at that 0 that we plugged in, so we know we have that local max at x = 0.

Now let's go ahead and plug 2 in here too. We have 6 * 2 - 1. That gives us a positive 6. Again, here, we just care about that sign here. This is a positive value.

And because this second derivative is positive here, that indicates to us that we have a local minimum here. So we have a local minimum at that value of x that we plugged in, which was x = 2. So we have located our local extrema here. We have a local max at x = 0 and a local min at x = 2. Now remember that in these problems, if you are asked to find the value of your max or min, you want to go ahead and plug your critical point back into your original function f(x).

So here, we would want to plug x = 0 and x = 2 back into that original function f(x) to really get those maximum and minimum values. Now, we're going to get more practice with this coming up next, and I'll see you there.

Use the second derivative test to find the local extrema of the given function.

$g(x)=x^3-6x^2+9x+2$

• A. Local max of $6$ at $x=1$, local min of 2 at $x=3$
• B. Local min of $6$ at $x=1$, local max of 2 at $x=3$
• C. Local max of $6$ at $x=-1$ and $x=1$, local min of $-6$ at $x=-3$ and $x=3$
• D. No local extrema

Use the second derivative test to find the local extrema of the given function.

$f\left(x\right)=\frac{x^2-4}{x^2+1}$

• A. Local max of $10$ at $x=0$
• B. Local min of $-4$ at $x=0$
• C. Local min of $-4$ at $x=-2$ and Local max of $10$ at $x=2$
• D. No local extrema

Use the second derivative test to find the local extrema of the given function.

$f\left(x\right)=4\sin x\cos x$; $0$ < $x$ < $\pi$

• A. Local min of $-2$ at $x=\frac{\pi}{2}$, local max of $8$ at $x=\pi$
• B. Local max of $2$ at $x=\frac{\pi}{2}$, local min of $-2$ at $x=\pi$
• C. Local min of $-8$ at $x=\frac{\pi}{4}$, local max of $8$ at $x=\frac{3\pi}{4}$
• D. Local max of $2$ at $x=\frac{\pi}{4}$, local min of $-2$ at $x=\frac{3\pi}{4}$

The Second Derivative Test: Finding Local Extrema Example 1

In this problem, we're asked to find the local extrema of the function \( f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 \). Now we're going to go ahead and start here by trying to use the second derivative test because it may be a bit quicker this way. So let's go ahead and first find our first derivative and set that equal to 0. So, finding our first derivative here using the power rule, \( f'(x) \) is going to be equal to \( 4x^3 - 12x^2 + 12x - 4 \). Now if we go ahead and factor this a little bit, we can go ahead and pull a 4 out of every single one of these terms.

So that gives us here a \( 4(x^3 - 3x^2 + 3x - 1) \). Now looking at this, this actually perfectly factors down to \( 4(x - 1)^3 \). So that's going to make it a little bit easier to set this equal to 0. Now setting this equal to 0, that really just gives me \( x - 1 = 0 \) because the 4 and the cubed will cancel. So if I add 1 to both sides, I can see that \( x = 1 \) is my one value for which that derivative will be equal to 0.

Now here we don't have to worry about that first derivative not existing because this is a basic polynomial, so I'm good, at this point, to keep on using the second derivative test. Now that we have that critical point or that point where our derivative is equal to 0, we can go ahead and plug this value into our second derivative and take a look at whether it's positive or negative. So finding my second derivative here, that's \( f''(x) \). Since I already factored this down to \( 4(x - 1)^3 \), I can go ahead and take the derivative using this form of that first derivative. That gives me \( 12(x - 1)^2 \).

Now multiplying that by whatever is inside there using the chain rule, that would just be multiplying by 1, so that's not going to do anything. Now if I go ahead and plug this value 1 into my second derivative here, that's what I want to do in this step, \( f''(1) \) will be \( 12(1 - 1)^2 \). Now this actually gives me 0. So what should I do here? Well, this is not a negative or positive value and because here our second derivative is equal to 0, that means that our second derivative test isn't going to work here and we need to find our local extrema by using the first derivative test.

This doesn't give us any information on whether this point, \( x = 1 \), is a local max or local min. So we want to revert back to using our first derivative test here. The second derivative didn't tell us anything. But we could proceed in using our first derivative test here. Remember, we can set up a sign chart and look at our sign changes of that first derivative.

So I'm going to go ahead and mark a one on my intervals here. So from negative infinity to 1 and from 1 on to infinity. Now I want to choose a test value in each of these intervals to plug back into that first derivative here. I'm going to go ahead and choose 0 on this interval and 2 on this interval. Remember, we want to choose values that make it easy to plug back into that first derivative.

So plugging these into that first derivative, I'm finding here \( f'(0) \) and \( f'(2) \) and paying attention to the sign of them. Remember, the sign changes are what's going to tell me whether this point is a local max or a local min based on the first derivative test. So plugging a 0 back into that first derivative here, that's going to give me \( 4(0 - 1)^3 \). That ends up giving me a negative 4 because this value is negative, so \( f' \) here is negative. That tells me that my function there is going to be decreasing on that interval.

Then if I plug 2 in here, that's \( 4(2 - 1)^3 \). This gives me a positive 4. I don't actually care about that value, just the sign because this first derivative is positive. That tells me my function is increasing on this interval. Based on this derivative and the first derivative test, that tells me that this point \( x = 1 \) is going to be a local minimum.

So my function has a local mininum at \( x = 1 \). Now in order to actually find what this value is or what this minimum is, we can revert back to this step 3, which is the exact same final step of our first_derivative test, and plug this back into our original function. So \( f(1) \) here. Now if I plug 1 back into my original function, that's \( 1^4 - 4 \times 1^3 + 6 \times 1^2 - 4 \times 1 + 1 \). Now this all ends up working out to just be 0.

So now I know that my function has a local minimum value of 0 at \( x = 1 \). Remember that if you start off using your second derivative test, you can only continue on using it to find your local extrema if you're only working with values where your first derivative is equal to 0, so not where it doesn't exist, and you end up getting a positive or negative value for that second derivative when you plug that in. Now if you end up getting that your second derivative is equal to 0, like we did here, then you need to revert back to using your first_derivative test rather than the second. Thanks for watching, and I'll see you in the next one.