Question

given the function ( g(x)=8 x^{3}-12 x^{2}-48 x ), find the first derivative, ( g^{prime}(x) ).
( g^{prime}(x)= )
notice that ( g^{prime}(x)=0 ) when ( x = 2 ), that is, ( g^{prime}(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^{prime prime}(x) ).
( g^{prime prime}(x)= )
evaluate ( g^{prime prime}(2) ).
( g^{prime prime}(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

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

Step2: Find the second derivative

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

Step3: Evaluate \(g^{\prime\prime}(2)\)

Substitute \(x = 2\) into \(g^{\prime\prime}(x)\).
\(g^{\prime\prime}(2)=48\times2-24\)
\(g^{\prime\prime}(2)=96 - 24=72\)

Step4: Determine concavity

Since \(g^{\prime\prime}(2)=72>0\), by the second - derivative test for concavity, the graph of \(y = g(x)\) is concave up at \(x = 2\).

Step5: Determine local extrema

By the second - derivative test for local extrema: if \(g^\prime(c)=0\) and \(g^{\prime\prime}(c)>0\), then \(g(x)\) has a local minimum at \(x = c\). Here \(c = 2\), \(g^\prime(2) = 0\) and \(g^{\prime\prime}(2)>0\), so \(g(x)\) has a local minimum at \(x = 2\).

Answer:

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