sketch the graph of the following function. list the coordinates of where extrema or points of inflection…

sketch the graph of the following function. list the coordinates of where extrema or points of inflection occur. state where the function is increasing or decreasing as well as where it is concave up or concave down. f(x)= -x³ + 9x² - 52 d. f is never concave up. f is concave down on (-∞,3) and (3,∞). e. f is concave up on (3,∞). f is concave down on (-∞,3). f. f is concave up on (-∞,3) and (3,∞). f is never concave down. choose the correct graph below.

sketch the graph of the following function. list the coordinates of where extrema or points of inflection occur. state where the function is increasing or decreasing as well as where it is concave up or concave down. f(x)= -x³ + 9x² - 52 d. f is never concave up. f is concave down on (-∞,3) and (3,∞). e. f is concave up on (3,∞). f is concave down on (-∞,3). f. f is concave up on (-∞,3) and (3,∞). f is never concave down. choose the correct graph below.

Answer

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=-x^{3}+9x^{2}-52$ using the power rule. $f'(x)=-3x^{2}+18x=-3x(x - 6)$.

Step2: Find critical points

Set $f'(x) = 0$. So, $-3x(x - 6)=0$. The critical points are $x = 0$ and $x = 6$.

Step3: Determine intervals of increase and decrease

Test intervals: For $x<0$, let $x=-1$, then $f'(-1)=-3\times(-1)\times(-1 - 6)=-21<0$, so $f(x)$ is decreasing on $(-\infty,0)$. For $0<x<6$, let $x = 1$, then $f'(1)=-3\times1\times(1 - 6)=15>0$, so $f(x)$ is increasing on $(0,6)$. For $x>6$, let $x = 7$, then $f'(7)=-3\times7\times(7 - 6)=-21<0$, so $f(x)$ is decreasing on $(6,\infty)$.

Step4: Find the second - derivative

Differentiate $f'(x)=-3x^{2}+18x$. $f''(x)=-6x + 18=-6(x - 3)$.

Step5: Find inflection point

Set $f''(x)=0$. So, $-6(x - 3)=0$, and the inflection point is at $x = 3$.

Step6: Determine concavity

Test intervals: For $x<3$, let $x = 2$, then $f''(2)=-6\times(2 - 3)=6>0$, so $f(x)$ is concave up on $(-\infty,3)$. For $x>3$, let $x = 4$, then $f''(4)=-6\times(4 - 3)=-6<0$, so $f(x)$ is concave down on $(3,\infty)$.

Answer:

E. f is concave up on $(3,\infty)$. f is concave down on $(-\infty,3)$.