# Math Help - Find the maximum value and minimum value of the function.

1. ## Find the maximum value and minimum value of the function.

I have this problem from my Calculus 1 homework and I don't know if I'm doing it correctly.
f (x) = x^3 - 12x

For stationary points: f ' (x) = 0
f ' (x) = 3x^2 - 12 (solve for 0)

3x^2 = 12
x^2= 4
x= +2 and -2
(2, -16) and (-2, 16)

f '' (x) = 6x
f '' (2)= 6(2)
= 12

f '' (-2)= 6(-2)
= -12

Also, i am not sure what is meant by the closed interval (-4, 4),

2. ## Re: Find the maximum value and minimum value of the function.

evaluate the function at the two endpoints and at the stationary points inside the open interval. the largest(smalest) is the absolute maximum(minimum) of the function in the closed interval.

3. ## Re: Find the maximum value and minimum value of the function.

Originally Posted by calculus123
...
Also, i am not sure what is meant by the closed interval (-4, 4)...
A closed interval, denoted with brackets, such as $[a,b]$ means you include the end-points, i.e., $a\le x\le b$.

An open interval, denoted with parentheses, such as $(a,b)$ means you do not include the end-points, i.e., $a.

4. ## Re: Find the maximum value and minimum value of the function.

Originally Posted by MarkFL2
A closed interval, denoted with brackets, such as $[a,b]$ means you include the end-points, i.e., $a\le x\le b$.

An open interval, denoted with parentheses, such as $(a,b)$ means you do not include the end-points, i.e., $a.
So do I have to do anything with the points (-4,4) like plug it somewhere into the equation or am I done?

5. ## Re: Find the maximum value and minimum value of the function.

Yes, you need to evaluate the function at the end-points to see if either are greater or smaller than the function's values at the critical points in the interval.

From these 4 points, choose the smallest as your absolute minimum, and the largest as your absolute maximum.

6. ## Re: Find the maximum value and minimum value of the function.

How did I do this one?

f(x)= x³ + 4/x² + 7; x < 0

f'(x) = 3x² - 8x^-3

3x² - 8x^-3 = 0

x^5 = 8/3

x = (8/3)^(1/5) ≈ 1.21

Let's check what the slope is at x = -1

3 + 8 = 11

It is increasing over the entire interval.

f(-3): -27 + 4/9 + 7

The value at x = -3 is negative

and the value of f(-1) = -1 + 4 + 7 is positive.
Therefore, only one root.

bump.