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