Question

Let h(x) be a function whose second derivative is h 00(x) = x 3 − 4x 2 + 5x. h(x) has critical points at x = −1, x = 0, and x = 1. Which of these points are local maxima, which are local minima, and which can you not tell just by using the second derivative test?

Answer 1

Answer:

*The function has a minimum in x=-1

*The function has a maximum in x=1

*The second derivative is not enough to determine if the function has either a maximum or a minimum in x=0.

Step-by-step explanation:

1. Evaluate the second derivative in the first critical point x=-1:

$h"(x)=x^{3}-4x^{2}+5x$

$h"(-1)=(-1)^{3}-4(-1^{2})+5(-1)$

$h"(-1)=-1-4-5$

$h"(-1)=-10$

As the value is smaller than zero, the function has a minimum in x=-1

2. Evaluate the second derivative in the second critical point x=1

$h"(x)=x^{3}-4x^{2}+5x$

$h"(1)=(1)^{3}-4(1^{2})+5(1)$

$h"(1)=1-4+5$

$h"(1)=2$

As the value is larger than zero, the function has a maximum in x=1

3. Evaluate the second derivative in the third critical point x=0

$h"(x)=x^{3}-4x^{2}+5x$

$h"(0)=(0)^{3}-4(0^{2})+5(0)$

$h"(0)=0-0+0$

$h"(0)=0$

As the value is equal to zero, the second derivative is not enough to determine if the function has either a maximum or a minimum in x=0.