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.\nf(x)= -x³ + 9x² - 52\n\nd. f is never concave up. f is concave down on (-∞,3) and (3,∞).\ne. f is concave up on (3,∞). f is concave down on (-∞,3).\nf. f is concave up on (-∞,3) and (3,∞). f is never concave down.\nchoose the correct graph below.\na. graph icon\nb. graph icon\nc. graph icon\nd. graph icon

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.\nf(x)= -x³ + 9x² - 52\n\nd. f is never concave up. f is concave down on (-∞,3) and (3,∞).\ne. f is concave up on (3,∞). f is concave down on (-∞,3).\nf. f is concave up on (-∞,3) and (3,∞). f is never concave down.\nchoose the correct graph below.\na. graph icon\nb. graph icon\nc. graph icon\nd. graph icon

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 increasing and decreasing intervals

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

Step5: Find inflection point

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

Step6: Determine concavity

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 $(3,\infty)$.

The function has a local minimum at $x = 0$, $f(0)=-52$, so the local - minimum point is $(0,-52)$. It has a local maximum at $x = 6$, $f(6)=-6^{3}+9\times6^{2}-52=-216 + 324-52 = 56$, so the local - maximum point is $(6,56)$. The inflection point is at $(3,f(3))$, and $f(3)=-3^{3}+9\times3^{2}-52=-27 + 81-52 = 2$. So the inflection point is $(3,2)$.

The function is increasing on $(0,6)$ and decreasing on $(-\infty,0)\cup(6,\infty)$. It is concave up on $(-\infty,3)$ and concave down on $(3,\infty)$.

Answer:

The local - minimum point is $(0,-52)$, the local - maximum point is $(6,56)$, the inflection point is $(3,2)$. The function is increasing on $(0,6)$, decreasing on $(-\infty,0)\cup(6,\infty)$, concave up on $(-\infty,3)$ and concave down on $(3,\infty)$.