absolute max and min

[kickingtoad]

\(\displaystyle f(x)=x^{3}-28x\)

On each interval below, find the value of x where the absolute maximum and the absolute minimum of f(x) occur, if they exist. If the absolute maximum or minimum does not exist, write "DNE".

(-infinity, infinity)
absolute min= does not exist
absolute max = does not exist

[-4.infinity)
absolute min=
absolute max= does not exist

[-4,0]
absolute min=
absolute max=

[0,10]
absolute min=
absolute max=

What is the correct way to solve this? I found the derivative of f(x) and plugged in the interval values to find the critical numbers, but those numbers weren't correct.

 

[DrSteve]

This is a continuous function. Thus, if the interval is closed, you can substitute the critical numbers and the two endpoints of the interval into the original function to get the absolute max and min (you can do this for the last two intervals). Otherwise you need to check carefully where the function is increasing and decreasing.

 

[kickingtoad]

Okay, I'm lost on how to find the absolute max and min by using the critical points.

To find the critical points you set \(\displaystyle f'(x)=0\)

\(\displaystyle f'(x)=3x^{2}-28\)

\(\displaystyle 3x^{2}-28=0\)
\(\displaystyle x=\pm\sqrt\frac{28}{3}=3.055\)

So then I insert 3.055 into the original function
\(\displaystyle (3.055^{3})-28(3.055)=-57.03\)
Is this the min for one of the intervals?

 

[Deleted member 4993]

Okay, I'm lost on how to find the absolute max and min by using the critical points.

To find the critical points you set \(\displaystyle f'(x)=0\)

\(\displaystyle f'(x)=3x^{2}-28\)

\(\displaystyle 3x^{2}-28=0\)
\(\displaystyle x=\pm\sqrt\frac{28}{3}= \pm 3.055\)

So then I insert 3.055 into the original function
\(\displaystyle (3.055^{3})-28(3.055)=-57.03\)

You have used the positive root - you need to use the negative root also - that will give the other value to compare.
Is this the min for one of the intervals?

 

[lookagain]

Okay, I'm lost on how to find the absolute max and min by using the critical points.

To find the critical points you set \(\displaystyle f'(x)=0\)

\(\displaystyle f'(x)=3x^{2}-28\)

\(\displaystyle 3x^{2}-28=0\)

\(\displaystyle x=\pm\sqrt\frac{28}{3}=3.055\)

So then I insert 3.055 into the original function

\(\displaystyle (3.055)^{3} -28(3.055)=-57.03 \ . \ . \ . \ .\ . \ . \ kickingtoad, \ the \ parentheses \ go \ around \ the\) \(\displaystyle critical \ value \ wherever \ you \ insert \ it \ for \ x, \ such \ as \ for \ the \ x^3 \ part.\)


Is this the min for one of the intervals?