Question

find the exact location of the relative and absolute extrema of the function (g(x)=-3x^{3}+36x + 7) on the interval (-5,5). (enter dne for \does not exist\ if there are none.)
(g(x)) has an absolute minimum at ((x,y)=()
(g(x)) has an absolute maximum at ((x,y)=()
(g(x)) has a relative minimum at ((x,y)=()
(g(x)) has a relative maximum at ((x,y)=()

Explanation:

Step1: Find the derivative

Differentiate $g(x)=-3x^{3}+36x + 7$ using power - rule. $g'(x)=-9x^{2}+36$.

Step2: Set the derivative equal to zero

Solve $-9x^{2}+36 = 0$. First, factor out - 9: $-9(x^{2}-4)=0$, then $x^{2}-4=(x - 2)(x + 2)=0$. So $x=-2$ or $x = 2$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $g(x)$ at $x=-5,-2,2,5$.
$g(-5)=-3(-5)^{3}+36(-5)+7=-3(-125)-180 + 7=375-180 + 7=202$.
$g(-2)=-3(-2)^{3}+36(-2)+7=-3(-8)-72 + 7=24-72 + 7=-41$.
$g(2)=-3(2)^{3}+36(2)+7=-3(8)+72 + 7=-24+72 + 7=55$.
$g(5)=-3(5)^{3}+36(5)+7=-3(125)+180 + 7=-375+180 + 7=-188$.

Step4: Determine extrema

• Absolute minimum: The smallest value among $g(-5),g(-2),g(2),g(5)$ is $-188$ at $x = 5$. So the absolute minimum is at $(x,y)=(5,-188)$.
• Absolute maximum: The largest value among $g(-5),g(-2),g(2),g(5)$ is $202$ at $x=-5$. So the absolute maximum is at $(x,y)=(-5,202)$.
• Relative minimum: Since $g'(x)$ changes sign from negative to positive at $x=-2$, the relative minimum is at $(x,y)=(-2,-41)$.
• Relative maximum: Since $g'(x)$ changes sign from positive to negative at $x = 2$, the relative maximum is at $(x,y)=(2,55)$.

Answer:

$g(x)$ has an absolute minimum at $(x,y)=(5,-188)$.
$g(x)$ has an absolute maximum at $(x,y)=(-5,202)$.
$g(x)$ has a relative minimum at $(x,y)=(-2,-41)$.
$g(x)$ has a relative maximum at $(x,y)=(2,55)$.