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 2 years ago
Calculus help! Find the extreme values of the function and where they occur. f(x)=x4sqrt(x)
 2 years ago
Calculus help! Find the extreme values of the function and where they occur. f(x)=x4sqrt(x)

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cxs379
 2 years ago
Best ResponseYou've already chosen the best response.1Find the derivative and set it equal to zero.

kumar2006
 2 years ago
Best ResponseYou've already chosen the best response.1The maximum and minimum are where y' = 0 By the product rule, if y = f(x) g(x), then y' = f(x) g'(x) + f'(x) g(x). So, y' = x^2 e^x + 2x e^x = 0; divide through by e^x, x^2 + 2x = 0; factoring, x (2 + x) = 0, so x = 0 and x = 2. Substitute into the original equation to get the y's, and we obtain (0,0) and (2, 0.54) as the extremes. Since the function always has a positive slope for x>0, then (0,0) must be a function minimum. Therefore (2, 0.54) is the function maximum.

dilfalicious
 2 years ago
Best ResponseYou've already chosen the best response.0so the derivative would be =12x^1/2?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0occurs at 12/sqrt(x) = 0 =>solve for x,

dilfalicious
 2 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=x4\sqrt{x}\] is the equation

brainshot3
 2 years ago
Best ResponseYou've already chosen the best response.0Looks like you are going to have a cusp because you are going to have an imaginary number as one of the values... No negatives in square roots.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0http://www4c.wolframalpha.com/Calculate/MSP/MSP3891a12dab5i5i5822400004i88c35ceba41di2?MSPStoreType=image/gif&s=12&w=320&h=130&cdf=RangeControl at the point where you see following shapedw:1334006278500:dw

cxs379
 2 years ago
Best ResponseYou've already chosen the best response.1\[f \prime(x)=12/\sqrt(x)\] \[1=2*\sqrt(x)\] \[\sqrt(x)=2\] x=4, critical number. Location is \[f(4)=(4)\]so (4, 4)

cxs379
 2 years ago
Best ResponseYou've already chosen the best response.1Its not 2*sqrt(x) its 2/sqrt(x)
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