Question

The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum value at what x value?
a. -1
b. 0
c. 1
d. 4/3
e. 5/3

Answer 1

B

Answer 2

can you explain it to me, please?

Answer 3

Hold on

Answer 4

find f''(x)

Answer 5

f''(x) = 4x^2 - 3x^3

what next?

Answer 6

Isn't that it

Answer 7

Ummm, looks like your \(f''(x)\) is a bit off.

Answer 8

The condition To attain Max is

f''(x) > 0

Answer 9

That is true

Answer 10

They want the max of the derivative.

Answer 11

So you want to find the critical numbers \(f''(x)=0\) or undefined. Then you'd plug them back into \(f'(x)\) to see where the max is.

Answer 12

nope @wio they ask..For where the max is attained

Answer 13

"The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum"

"f'(x) attains its maximum"

Answer 14

aha! f"(x) = 4x^2 - 4x^3.

i got my critical numbers. x = 0 & x = 1?

Answer 15

That looks good. So your local maximum value is at either x=0 or x=1. Also, since your first derivative is a negative quartic (has a \(-x^4\) term), you have a parabola type shape opening downwards, so the local maximum is the global maximum. Now you just need to determine which one is the maximum.

Answer 16

make sense so far?

Answer 17

i think. i got f'(0) = 0 and f'(1) = 1/3. so the answer is... 1?

Answer 18

Bingo.

Answer 19

All riiiiiiiiight. thank you. and wio. but s/he left.

Answer 20

You're welcome.