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dilfalicious
Calculus help! Find the extreme values of the function and where they occur. f(x)=x-4sqrt(x)
Find the derivative and set it equal to zero.
The 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.
so the derivative would be =1-2x^-1/2?
occurs at 1-2/sqrt(x) = 0 =>solve for x,
\[f(x)=x-4\sqrt{x}\] is the equation
Looks 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.
http://www4c.wolframalpha.com/Calculate/MSP/MSP3891a12dab5i5i5822400004i88c35ceba41di2?MSPStoreType=image/gif&s=12&w=320&h=130&cdf=RangeControl at the point where you see following shape|dw:1334006278500:dw|
\[f \prime(x)=1-2/\sqrt(x)\] \[1=2*\sqrt(x)\] \[\sqrt(x)=2\] x=4, critical number. Location is \[f(4)=(-4)\]so (4, -4)
Its not 2*sqrt(x) its 2/sqrt(x)