How do use the first derivative test to determine the local extrema #f(x) = xsqrt(25-x^2)#?

1 Answer
Oct 14, 2015

Answer:

The extrema of your functions are #\pm 5/sqrt(2)#.

Explanation:

You compute the first derivative and find its roots. The derivative is computed as follows:

#D(x*sqrt(25-x^2))=D(x)*sqrt(25-x^2) + x*D(sqrt(25-x^2))#.

#D(x)# is obviously one, while

#D(sqrt(25-x^2))=D(25-x^2)^{1/2}=1/2(25-x^2)^{-1/2}*D(25-x^2)#

Since #D(25-x^2)=-2x#, the whole expression becomes

#f'(x)=sqrt(25-x^2) + x*{-cancel(2)x}/{cancel(2)sqrt(25-x^2)}#

#=sqrt(25-x^2)-{x^2}/{sqrt(25-x^2)}#

#={(25-x^2)-x^2}/{sqrt(25-x^2)}={25-2x^2}/{sqrt(25-x^2)}#.

This function equals zero if and only if its numerator equals zero, so

#f'(x)=0 \iff 2x^2=25 \iff x^2=25/2 \iff x=\pm 5/sqrt(2)#.

These are the local extrema of your function.