locate the critical points of the following function. then use the second derivative test to determine…

locate the critical points of the following function. then use the second derivative test to determine whether they correspond to local maxima, local minima, or neither. f(x)=x^3 - 9x^2 what is(are) the critical point(s) of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) is(are) x = 0,6. (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. there are no critical points for f. find f(x). f(x)=6x - 18 what is/are the local maximum/maxima of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the local maximum/maxima of f is/are at x = . (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. there is no local maximum of f.

locate the critical points of the following function. then use the second derivative test to determine whether they correspond to local maxima, local minima, or neither. f(x)=x^3 - 9x^2 what is(are) the critical point(s) of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) is(are) x = 0,6. (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. there are no critical points for f. find f(x). f(x)=6x - 18 what is/are the local maximum/maxima of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the local maximum/maxima of f is/are at x = . (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. there is no local maximum of f.

Answer

Explanation:

Step1: Recall critical - point definition

Critical points occur where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined. First, find the first - derivative of $f(x)=x^{3}-9x^{2}$. Using the power rule $\frac{d}{dx}(x^{n}) = nx^{n - 1}$, we have $f^{\prime}(x)=3x^{2}-18x$. Set $f^{\prime}(x)=0$: [3x^{2}-18x = 3x(x - 6)=0] Solving $3x(x - 6)=0$ gives $x = 0$ and $x = 6$.

Step2: Recall second - derivative test

The second - derivative is $f^{\prime\prime}(x)=6x - 18$. Evaluate $f^{\prime\prime}(x)$ at the critical points: When $x = 0$, $f^{\prime\prime}(0)=6\times0-18=-18<0$. By the second - derivative test, if $f^{\prime}(c)=0$ and $f^{\prime\prime}(c)<0$, then $f(x)$ has a local maximum at $x = c$. So, $f(x)$ has a local maximum at $x = 0$. When $x = 6$, $f^{\prime\prime}(6)=6\times6-18=36 - 18 = 18>0$. By the second - derivative test, if $f^{\prime}(c)=0$ and $f^{\prime\prime}(c)>0$, then $f(x)$ has a local minimum at $x = c$.

Answer:

A. The local maximum/maxima of $f$ is/are at $x = 0$