given the function $g(x)=8x^{3}-12x^{2}-48x$, find the first derivative, $g(x)$.\n$g(x)=$\nnotice that…

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

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

Answer

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.