# Thread: absolute max/ absolute min

1. ## absolute max/ absolute min

Consider the function f(x) = x4 − 72 x2 + 5, −5 ≤ x ≤ 13. This function has an absolute minimum value equal to __
and an absolute maximum value equal to ____

what i did is this:

f'(x) = 4(x^3)-144x
0=4(x^3)- 144x
0= 4x [ x^2 - 36 ]
36=x^2
x = 6

is this right so far?

now to check the endpoints ....

i have
-5
6
and
13

that's all i have to check right?

2. ## Re: absolute max/ absolute min

Originally Posted by habibixox
Consider the function f(x) = x4 − 72 x2 + 5, −5 ≤ x ≤ 13. This function has an absolute minimum value equal to __
and an absolute maximum value equal to ____

what i did is this:

f'(x) = 4(x^3)-144x
0=4(x^3)- 144x
0= 4x [ x^2 - 36 ]
36=x^2
x = 6

is this right so far?

now to check the endpoints ....

i have
-5
6
and
13

that's all i have to check right?
You will find it easier for others to follow if you tell us in words what you are trying to do rather than just dump a load of manipulation on the page.

The global maximum (minimum) will be either a calculus type maximum (minimum) or it will occur at a boundary point. You have the derivative correct:

$f'(x)=4x^3-144x=4x(x^2-36)$

Now a calculus type critical point occurs where the derivative is zero so we look at the roots of:

$f'(x)=4x(x^2-36)=0$

which are $x=0$, and $x=\pm 6$, but we drop the $x=-6$ as it is outside the interval of interest.

So now you need to examine the value of the function at $x=-5$, $x=0$, $x=6$ and $x=13$.

(By the way the question is asking for the maximum (minimum) of the function not the value of $x$ where it attains the maximum (minimum) )

CB