Use the first derivative test to determine the location of each local extremum and the value of the function at thatextremum.
-$f(x)=\frac{x^{2}-8 x+16}{x-5}$
A) No local extrema
B) Local maximum at $(4,0)$; local minimum at $(6,4)$
C) Local maximum at $(0,16)$; local minima at $(4,0)$ and $(14,11.11)$
D) Local maximum at $(6,4)$; local minimum at $(4,0)$

Question

Quizplus · Accepted Answer

The first derivative is $f'(x)=\frac{2(x-4)}{(x-5)^2}$. The critical numbers are $x=4$ and $x=5$. The first derivative is positive on the interval $(-\infty,4)$ and negative on the interval $(4,\infty)$, so $x=4$ is a local maximum. The first derivative is negative on the interval $(-\infty,5)$ and positive on the interval $(5,\infty)$, so $x=5$ is a local minimum. The value of the function at the local maximum is $f(4)=0$, and the value of the function at the local minimum is $f(6)=4$.

Use the First Derivative Test to Determine the Location of Each