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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 6

Lesson 7: Justifying properties of functions using the second derivative

# Second derivative test

Sal justifies the second derivative test, which is a way of determining relative minima & maxima, and gives an example.

## Want to join the conversation?

• The second derivative is is negative at c, so we conclude that the slope is decreasing around c?
How can we conclude that when we only know the the second derivative is only negative at c and only c and not at points around c? • I'm not sure how to prove this to you, but I think I can give you an intuition for why this is true.

From earlier videos we know that a derivative at c, means that the function is continuous at c. This means that there must be some values for x very close to c where `f(x)` is defined and continuous - I'm going to call this "window" `c ± ∆x`.
Now imagine that we make a very tiny (infinitesimal) step in the negative direction to `c-∆x`. Will `f(c-∆x)` be less than, greater than or equal to `f(c)`?

I hope you can see that if `∆x` gets small enough, then `f(c-∆x) < f(c)` because otherwise you would break the line and then the function wouldn't be continuous at x=c, which would be a contradiction!

The same argument holds for `f(c+∆x)`.

Consequently, the output for the function `f(x)` must drop at least a tiny amount as you move either direction from `x=c`, and thus by definition you have a local maximum.
• What about if the second derivative is equal to zero? Is it always inconclusive? • What happens if f prime does not exist around an x value c? Why would the test not apply? • Why f"(c)=0 inconclusive? • What does f''(x)=0 look like on a graph? I've gotten asked this on my homework and have no idea how to interpret it. • at , how do we know that the second derivative is less than zero? and how do we even know if this function can have a second derivative? what is needed to, what defines a second derivative? thank you! • The second derivative is the derivative of the first derivative.
e.g.`f(x) = x³ - x²f'(x) = 3x² - 2xf"(x) = 6x - 2`

So, to know the value of the second derivative at a point (`x=c`, `y=f(c)`) you:
1) determine the first and then second derivatives
2) solve for f"(c)

e.g. for the equation I gave above f'(x) = 0 at x = 0, so this is a critical point.`f"(0) = 6•0 - 2 = -2`
Therefore, f(x) is concave downward at x=0 and this critical point is a local maximum.

Can you do the same for the other critical point?
• Is Second derivarive for the maximum and minimum of the curve , right ?
(1 vote) • No. To find the maximum and minimum points, you use the first derivative. To get a max or min, the points you want to consider are where the function stops increasing and begins to decrease, or stops decreasing and begins to increase. If the function is continuous, there will be a point there where the function's rate of change (derivative) is zero. At this point, the increasing function stops increasing, creating a potential maximum point.
• Would love to see visual examples of how h'(x)=0 and h''(x)=0 could be a max, min or nothing. • The functions x^3, -x^4, and x^4 all have the property that their first and second derivatives are zero at x = 0. Yet the first function has no min or max at all, the second function has a local (and also global) max at x = 0, and the third function has a local (and also global) min at x = 0.

If you want visual examples, you can graph these three functions manually or on a graphing calculator.

Have a blessed, wonderful day!
• when can the 2nd derivative test not be used? • If f(x)=x^3 for closed function of [-2,2], first derivative is 3x^2, so critical points are -2, 0, 2. Now, second derivative = 6x. Therefore, f"(-2)=-12 and f"(2)=12 So, -2 must be the max value and 2 must be the min value. But f(-2)=-8 which is lesser than f(2)=8. Can someone explain where I went wrong?
(1 vote) • Because you're not looking at the entire domain of the function (this function's domain is across all real numbers), you need to be careful on what critical points actually are.

So, you've correctly gotten that the critical points are -2, 0, and 2. This is because you're always interested what happens at the end of the interval (giving you -2 and 2) and you're also interested in what happens if the deriviative is 0 (giving you 0).

Now, the second derivate test only applies if the derivative is 0. This means, the second derivative test applies only for x=0. At that point, the second derivative is 0, meaning that the test is inconclusive. So you fall back onto your first derivative. It is positive before, and positive after x=0. Therefore, x=0 is an inflection point.

Now, we go back to x=-2 and x=2. We can't use derivatives here because they lie at the ends of the interval we are looking at. So, we simply evaluate the function. And we see that x=-2 is a minimum of that function and x=2 is a maximum of that function on that interval.