Question

Find the local min and max values of f.
a) f(x) = 2x^2/3(3-4x^1/3)
(I'm just not sure how to take the derivative of it, please help.) Thanks!

Answer 1

Oh oops, sorry, I wrote the function wrong, it's 2x^2/3...

Answer 2

well that makes it a bit trickier, eh? But only slightly so fortunately. If we start with the distributing

\[f(x) = 2x^{\frac{2}{3}}(3)-(2x^{\frac{2}{3}})(4x^{\frac{1}{3}})\]
Combine like terms on the right term, then take the derivative of each separately.

Answer 3

okay, I got 4(x^-1/3-2)?

Answer 4

Yup yup yup ^_^

Answer 5

How would I isolate the x?

Answer 6

cube both sides?

Answer 7

I'm also an idiot and wrote out the general form of the derivative up top completely wrong, so apologies there first.

And yeah, pretty much. The cube root is on the bottom, so you'll have to cross multiply them to make it positive.

Answer 8

so it would be x= 1/8?

Answer 9

Methinks yes. There might also be an imaginary solution too since you loose information in the cubing of it, but I'm not 100% sure...

Answer 10

At the back of book, it says
f(1/8)=1/2 (local max)
f(0)=0 (local min)
I'm not sure where the zero is coming from? And also didn't quite undertstand how to determine local max and min.

Answer 11

Do this is the graph, so visualize it.

To find out if something is a local min/max, you have to test point around the root. So if you test f'(x=1/9) - a value smaller than 1/8 - and f'(x=1/7) - larger - then you can "see" if the derivative is positive or negative. If the number is positive that means the graph is increasing at that point with the opposite also being true, and testing if before/after the f'(x) = 0 point the slope is increasing/decreasing let's you see if it's a hill or a hole. In this example, the slope before the f'(x) = 0 is positive, and after it is negative, so we know that the x=1/8 root is a max.


Phew.

For the x=0, you can see on the graph it's a low point, but I'm spacing at the moment how you can test for it....

Sorry for the wait. I was trying to remember but am just drawing blanks

Answer 12

~So this is the graph, to visualize it

Answer 13

OH! I think I remember. The function f'(x) is undefined at x=0 because you're dividing by zero, so if you test around like with the other root then you can see what's happening around x=0, and it turns out to be a root.

Answer 14

~it turns out to be a minimum

My brain is not happy with me tonight.....

Answer 15

Oh, I see! Thank you for explaining it to me and the graph! Best person ever! ^_^

Answer 16

Very welcome. Glad I could help, and apologies for taking so long :P

Answer 17

Well, good night! ^_^

Answer 18

g'night ^_^