## jesusisabiscuit 2 years ago 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

1. Lilmoney

B

2. jesusisabiscuit

can you explain it to me, please?

3. Lilmoney

Hold on

4. Yahoo!

find f''(x)

5. jesusisabiscuit

f''(x) = 4x^2 - 3x^3 what next?

6. Lilmoney

Isn't that it

7. wio

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

8. Yahoo!

The condition To attain Max is f''(x) > 0

9. Lilmoney

That is true

10. wio

They want the max of the derivative.

11. wio

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.

12. Yahoo!

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

13. wio

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

14. jesusisabiscuit

aha! f"(x) = 4x^2 - 4x^3. i got my critical numbers. x = 0 & x = 1?

15. KingGeorge

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.

16. KingGeorge

make sense so far?

17. jesusisabiscuit

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

18. KingGeorge

Bingo.

19. jesusisabiscuit

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

20. KingGeorge

You're welcome.