Using the Second Derivative Test to Determine Extrema

#5.7 Using the Second Derivative Test to Determine Extrema

You’ve probably noticed by now that Unit 5 deals with analytical applications of differentiation; that means that a function’s derivatives can tell us something about its behaviors. We learned from 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema that the first derivative of a function tells us information on where its relative extrema (aka local maximum and minimum) will be.

What does the second derivative tell us, then? Answer: Whether a critical point we found yields a local maximum or minimum (it can’t be both)! 🧠

#✏️ Warm-up: Finding Critical Points

Let’s do a quick recap of critical points from key topic 5.2! To refresh, a critical point is a point on a function where…

• the first derivative equals zero or
• fails to exist Analytically, the first bullet point is easier to wrap your head around. Let’s take a look at a quick example:

$$f(x)=\frac{2}{3}x^3-\frac{5}{2}x^2-3x$$

Computing the first derivative gives us:

$$f'(x)=2x^{2}-5x-3$$

To find where the first derivative equals zero, we can factor and then set f’(x) = 0:

$$0=2x^2-5x-3$$

$$0=(2x+1)(x-3)$$

Looks familiar, right? We proceed as if we’re finding the roots (x) in a regular equation. In this case, the two x values are our critical points!

$$x=-\frac{1}{2}, x=3$$

Not bad, right? What do we do with these critical point values, then?

We learned two things from 5.6 Determining Concavity of Functions over Their Domains about what the second derivative of a function can tell us information of:

1. 🌈 The function’s intervals of upward or downward concavity.
2. $f’’(x)$ is positive (or greater than zero, aka $f’’(x)>0$) = concave up
3. $f’’(x)$ is negative (or less than zero, aka $f’’(x)<0$) = concave down
4. 🪝 The function graph’s inflection points (where concavity changes).

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#🥈 Extending the Second Derivative Test

Now, let’s connect these ideas to the critical points we mentioned earlier: by knowing the concavities before and after the critical points, we can determine where our local minima and maxima are! 🗺️

#🪜 Second Derivative Test Steps

Here are some steps that we’ll go through:

1. Find the critical points of $f(x)$ using $f'(x)$.
2. Plug the critical points into $f''(x)$ to get a y-value.
3. Determine whether you have a local min or max!

#📝 Second Derivative Test Walkthrough

To illustrate, let’s continue with our example while we walk through the steps pertaining to the Second Derivative Test.

#🥇 Step 1: Find f(x)’s critical points using f’(x)

Done! We found our critical points: $x = -0.5$ and $x = 3$ earlier.

#🥈 Step 2: Plug critical points into f’’(x) to get a y-value.

From above, we found $f’(x)$ to be:

$$f'(x)=2x^2-5x-3$$

Calculating the second derivative:

$$f''(x)=4x-5$$

Then, we plug our critical points:

$$f''(-\frac{1}{2})=4(-\frac{1}{2})-5=-7$$

$$f''(3)=4(3)-5=7$$

These numbers give us the y-values for our local extrema and thus the whole coordinates: $(-0.5, -7)$ and $(3, 7)$. We’re ready for our final step…

#🥉 Step 3: Determine whether you have a local min or max!

In our case, $(-0.5, -7)$ is a local maximum and $(3, 7)$ is a local minimum. This is because of the work we did in step 2!

Conceptually, why does this make sense? Tap into your imagination… imagine a hill and a bowl!

• 🗻 If a point represents a maximum and is situated at the peak of a hill, the point appears to lie on the concave-down surface of said hill.
• 🥣 Conversely, if a point signifies a minimum and is located at the bottom of an upright bowl, it suggests that the point lies on the concave-up surface of the bowl. Therefore, when the function f is concave down at a critical point (we’ll call it “c”), it implies a maximum, whereas f being concave up at “c” indicates a minimum.

Feel more confident? Try the following examples on your own before looking at the solutions to see which areas you should do more practice on! 📣

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#💭 Second Derivative Test: Let’s Practice!

For each of the following functions f(x), determine whether each of their critical points are local maxima or minima.

#Second Derivative Test: Example 1

$$f(x)=4\sin(x), 0<x<2\pi$$

Let’s find our first and second derivatives:

$$f'(x)=4\cos(x)$$

$$f''(x)=-4\sin(x)$$

Setting $f’(x) = 0$ to find critical points x:

$$0=4\cos(x)$$

$$0=\cos(x)$$

Thinking back to the Unit Circle from trigonometry, at which angles $x$ will $cos(x) = 0$? We’ll get our critical points:

$$x=\frac{\pi}{2}, x=\frac{3\pi}{2}$$

Plugging these values into our second derivative:

$$f''(\frac{\pi}{2})=-4\sin(\frac{\pi}{2})=-4*1=-4$$

$$f''(\frac{3\pi}{2})=-4\sin(\frac{3\pi}{2})=-4*-1=4$$

Since $-4 < 0$, $(\frac{\pi}{2}, -4)$ is a local maximum of $f(x)$. On the other hand, since $4 > 0$, $(\frac{3\pi}{2}, 4)$ is a local minimum of f(x). We’re done! ⭐

#Second Derivative Test: Example 2

$$f(x)=247ln(x^2)$$

Calculating f’(x) and f’’(x) gives us:

$$f'(x)=\frac{247}{x}$$

$$f''(x)=-\frac{247}{x^2}$$

Setting f’(x) = 0…

$$0=\frac{247}{x}$$

Wait a second. This doesn’t give us a solution. What do we do in this case, then? This brings up another element of the Second Derivative Test:

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#💫 Closing

Great work! Now we applied the second derivative test to different functions to determine extrema. There’s one key point that we want to leave you with…

When dealing with a continuous function that has just one critical point across its entire domain and this particular point serves as a local extremum within a specific interval, it also stands as the global extremum for the function within that particular interval.

This is because there are no other critical points to challenge its extremum status within that specific interval. This is due to the function having only one peak or valley (extremum) within that interval. 🗻