Use the first derivative test to determine the location of each local extremum and the value of the function at thatextremum.
-$f(x)=\frac{x^{2}}{x^{2}+5}$
A) Local maximum at $(0,0)$
B) No local extrema
C) Local minimum at $(0,0)$
D) Local minimum at $(5,0.83333333)$

Question

Quizplus · Accepted Answer

The first derivative is $f'(x)=\frac{2x(x^2-5)}{(x^2+5)^2}$. The only critical number is $x=0$. Since $f'(x)>0$ for $x>0$ and $f'(x)<0$ for $x<0$, $f(x)$ has a local minimum at $x=0$.

Use the First Derivative Test to Determine the Location of Each