Calculus help! Find the extreme values of the function and where they occur. f(x)=x-4sqrt(x)

Question

anonymous · Answer

Find the derivative and set it equal to zero.

anonymous · Answer

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.

anonymous · Answer

so the derivative would be =1-2x^-1/2?

experimentx · Answer

occurs at 1-2/sqrt(x) = 0 =>solve for x,

anonymous · Answer

$$f(x)=x-4\sqrt{x}$$ is the equation

anonymous · Answer

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.

experimentx · Answer

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|