Question

given the function $g(x)=8x^{3}-12x^{2}-48x$, find the first derivative, $g(x)$.
$g(x)=$
notice that $g(x)=0$ when $x = 2$, that is, $g(2)=0$.
now, we want to know whether there is a local minimum or local maximum at $x = 2$, so we will use the second derivative test.
find the second derivative, $g(x)$.
$g(x)=$
evaluate $g(2)$.
$g(2)=$
based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x = 2$?
answer either up or down -- watch your spelling!!
at $x = 2$ the graph of $g(x)$ is concave
based on the concavity of $g(x)$ at $x = 2$, does this mean that there is a local minimum or local maximum at $x = 2$?
answer either minimum or maximum -- watch your spelling!!
at $x = 2$ there is a local

Explanation:

Step1: Find the first derivative \(g'(x)\)

Use the power rule \((x^n)' = nx^{n - 1}\).
For \(g(x)=8x^{3}-12x^{2}-48x\), we have:
\(g'(x)=(8x^{3})'-(12x^{2})'-(48x)'\)
\(g'(x)=8\times3x^{2}-12\times2x - 48\)
\(g'(x)=24x^{2}-24x - 48\)

Step2: Find the second derivative \(g''(x)\)

Differentiate \(g'(x)=24x^{2}-24x - 48\) using the power rule.
\(g''(x)=(24x^{2})'-(24x)'-(48)'\)
\(g''(x)=24\times2x-24\)
\(g''(x)=48x - 24\)

Step3: Evaluate \(g''(2)\)

Substitute \(x = 2\) into \(g''(x)\):
\(g''(2)=48\times2-24\)
\(g''(2)=96 - 24\)
\(g''(2)=72\)

Step4: Determine concavity

If \(g''(a)>0\), the graph of \(y = g(x)\) is concave up at \(x = a\). Since \(g''(2)=72>0\), at \(x = 2\) the graph of \(g(x)\) is concave up.

Step5: Determine local minimum or maximum

By the second - derivative test, if \(g'(c) = 0\) and \(g''(c)>0\), then \(g(x)\) has a local minimum at \(x = c\). Since \(g'(2)=0\) and \(g''(2)=72>0\), at \(x = 2\) there is a local minimum.

Answer:

\(g'(x)=24x^{2}-24x - 48\)
\(g''(x)=48x - 24\)
\(g''(2)=72\)
At \(x = 2\) the graph of \(g(x)\) is concave up.
At \(x = 2\) there is a local minimum.