Question

given the function $g(x) = 6x^3 + 9x^2 - 216x$, find the first derivative, $g(x)$.
$g(x) = \square$

notice that $g(x) = 0$ when $x = 3$, that is, $g(3) = 0$.

now, we want to know whether there is a local minimum or local maximum at $x = 3$, so we will use the second derivative test.
find the second derivative, $g(x)$.
$g(x) = \square$

evaluate $g(3)$.
$g(3) = \square$

based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x = 3$
at $x = 3$ the graph of $g(x)$ is select an answer

based on the concavity of $g(x)$ at $x = 3$, does this mean that there is a local minimum or local maximum at $x = 3$
at $x = 3$ there is a local select an answer

question help: video message instructor

Explanation:

Step1: Find first derivative

Use power rule: \((x^n)' = nx^{n - 1}\). For \(g(x)=6x^3 + 9x^2 - 216x\),

• Derivative of \(6x^3\): \(6\times3x^{2}=18x^{2}\)
• Derivative of \(9x^2\): \(9\times2x = 18x\)
• Derivative of \(-216x\): \(-216\)

So \(g'(x)=18x^{2}+18x - 216\)

Step2: Find second derivative

Differentiate \(g'(x)=18x^{2}+18x - 216\) using power rule:

• Derivative of \(18x^2\): \(18\times2x = 36x\)
• Derivative of \(18x\): \(18\)
• Derivative of \(-216\): \(0\)

So \(g''(x)=36x + 18\)

Step3: Evaluate \(g''(3)\)

Substitute \(x = 3\) into \(g''(x)\):
\(g''(3)=36\times3 + 18 = 108 + 18 = 126\)

Step4: Concavity at \(x = 3\)

Since \(g''(3)=126>0\), the graph is concave up at \(x = 3\) (positive second derivative means concave up).

Step5: Local min/max at \(x = 3\)

For second derivative test: if \(g'(c)=0\) and \(g''(c)>0\), then \(x = c\) is a local minimum. Since \(g'(3)=0\) and \(g''(3)>0\), there is a local minimum at \(x = 3\).

Answer:

s:

• \(g'(x)=\boldsymbol{18x^{2}+18x - 216}\)
• \(g''(x)=\boldsymbol{36x + 18}\)
• \(g''(3)=\boldsymbol{126}\)
• At \(x = 3\) the graph of \(g(x)\) is \(\boldsymbol{\text{concave up}}\)
• At \(x = 3\) there is a local \(\boldsymbol{\text{minimum}}\)