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

#### calculus123

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),

#### hedi

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.

1 person

#### MarkFL

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

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

1 person

#### calculus123

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

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

#### MarkFL

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.

1 person

#### calculus123

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.