Use the first derivative test to determine the location of each local extremum and the value of the function at thatextremum.
-$f(x)=(4-2 x)^{3 / 5}-4$
A) No local extrema
B) Local maximum at $(2,-4)$
C) Local minimum at $(2,-4)$
D) Local minimum at $(0,-4)$

Question

Quizplus · Accepted Answer

The first derivative is $f'(x)=-\frac{12}{5}(4-2x)^{2/5}$. This is negative for $x>2$ and positive for $x<2$, so there is a local minimum at $x=2$. The value of the function at this extremum is $f(2)=(4-2(2))^{3/5}-4=-4$.

Use the First Derivative Test to Determine the Location of Each