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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

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can you explain it to me, please?
Hold on

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Other answers:

find f''(x)
f''(x) = 4x^2 - 3x^3 what next?
Isn't that it
Ummm, looks like your \(f''(x)\) is a bit off.
The condition To attain Max is f''(x) > 0
That is true
They want the max of the derivative.
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.
nope @wio they ask..For where the max is attained
"The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum" "f'(x) attains its maximum"
aha! f"(x) = 4x^2 - 4x^3. i got my critical numbers. x = 0 & x = 1?
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.
make sense so far?
i think. i got f'(0) = 0 and f'(1) = 1/3. so the answer is... 1?
All riiiiiiiiight. thank you. and wio. but s/he left.
You're welcome.

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