Question

answer the following questions about the function whose derivative is $f(x)=(x - 6)^2(x + 7)$. a. what are the critical points of $f$? b. on what open intervals is $f$ increasing or decreasing? c. at what points, if any, does $f$ assume local maximum and minimum values? (simplify your answer. use a comma to separate answers as needed.) b. the function $f$ has no critical points. b. determine where $f$ is increasing and decreasing. select the correct choice below and fill in the answer box to complete your choice. (type your answer in interval notation. use a comma to separate answers as needed.) a. the function is increasing on the open interval(s) $(-7,6),(6,infty)$, and decreasing on the open interval(s) $(-infty,-7)$. b. the function $f$ is decreasing on the open interval(s) , and never increasing. c. the function $f$ is increasing on the open interval(s) , and never decreasing

Explanation:

Step1: Find critical points

Critical points occur where \(f^{\prime}(x) = 0\).
Set \((x - 6)^{2}(x + 7)=0\).
Using the zero - product property:
If \((x - 6)^{2}=0\), then \(x = 6\).
If \(x+7 = 0\), then \(x=-7\).

Step2: Analyze the sign of \(f^{\prime}(x)\)

• For \(x<-7\):

Let \(x=-8\), then \(f^{\prime}(-8)=(-8 - 6)^{2}(-8 + 7)=(-14)^{2}\times(-1)=196\times(-1)<0\).

• For \(-7:

Let \(x = 0\), then \(f^{\prime}(0)=(0 - 6)^{2}(0 + 7)=36\times7>0\).

• For \(x>6\):

Let \(x = 7\), then \(f^{\prime}(7)=(7 - 6)^{2}(7 + 7)=1\times14>0\).

Step3: Determine local maxima and minima

Since \(f^{\prime}(x)\) changes sign from negative to positive at \(x=-7\), by the first - derivative test, \(f(x)\) has a local minimum at \(x=-7\).
Since \(f^{\prime}(x)\) does not change sign at \(x = 6\) (the sign of \(f^{\prime}(x)\) is positive on both sides of \(x = 6\)), \(x = 6\) is not a local maximum or minimum.

Answer:

a. The critical points of \(f\) are \(x=-7\) and \(x = 6\).
b. The function is increasing on the open intervals \((-7,6)\cup(6,\infty)\) and decreasing on the open interval \((-\infty,-7)\).
c. The function \(f\) has a local minimum at \(x=-7\) and no local maximum.