Question

given the function $g(x)=6x^{3}+9x^{2}-360x$, find the first derivative, $g(x)$. notice that $g(x)=0$ when $x = 4$, that is, $g(4)=0$. now, we want to know whether there is a local minimum or local maximum at $x = 4$, so we will use the second derivative test. find the second derivative, $g(x)$. evaluate $g(4)$. based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x = 4$? answer either up or down -- watch your spelling!! at $x = 4$ the graph of $g(x)$ is concave. based on the concavity of $g(x)$ at $x = 4$, does this mean that there is a local minimum or local maximum at $x = 4$? answer either minimum or maximum -- watch your spelling!! at $x = 4$ there is a local. question help: video

Explanation:

Step1: Find the first - derivative

Use the power rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$.
$g(x)=6x^{3}+9x^{2}-360x$
$g^{\prime}(x)=3\times6x^{2}+2\times9x - 360=18x^{2}+18x - 360$

Step2: Find the second - derivative

Differentiate $g^{\prime}(x)$ using the power rule.
$g^{\prime}(x)=18x^{2}+18x - 360$
$g^{\prime\prime}(x)=2\times18x+18=36x + 18$

Step3: Evaluate $g^{\prime\prime}(4)$

Substitute $x = 4$ into $g^{\prime\prime}(x)$.
$g^{\prime\prime}(4)=36\times4+18=144 + 18=162$

Step4: Determine concavity

If $g^{\prime\prime}(a)>0$, the graph of $y = g(x)$ is concave up at $x=a$. Since $g^{\prime\prime}(4)=162>0$, the graph of $g(x)$ is concave up at $x = 4$.

Step5: Determine local extrema

If $g^{\prime}(a)=0$ and $g^{\prime\prime}(a)>0$, then $g(x)$ has a local minimum at $x=a$. Since $g^{\prime}(4)=0$ and $g^{\prime\prime}(4)>0$, $g(x)$ has a local minimum at $x = 4$.

Answer:

$g^{\prime}(x)=18x^{2}+18x - 360$
$g^{\prime\prime}(x)=36x + 18$
$g^{\prime\prime}(4)=162$
up
minimum