given the function $g(x) = 6x^3 + 45x^2 + 72x$, find the first derivative, $g(x)$.
$g(x) =$

notice that $g(x) = 0$ when $x = -1$, that is, $g(-1) = 0$.

now, we want to know whether there is a local minimum or local maximum at $x = -1$, so we will use the second derivative test.
find the second derivative, $g(x)$.
$g(x) =$

evaluate $g(-1)$.
$g(-1) =$

based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x = -1$?
at $x = -1$ the graph of $g(x)$ is select an answer

based on the concavity of $g(x)$ at $x = -1$, does this mean that there is a local minimum or local maximum at $x = -1$?
at $x = -1$ there is a local select an answer

Question

given the function $g(x) = 6x^3 + 45x^2 + 72x$, find the first derivative, $g(x)$.
$g(x) =$

notice that $g(x) = 0$ when $x = -1$, that is, $g(-1) = 0$.

now, we want to know whether there is a local minimum or local maximum at $x = -1$, so we will use the second derivative test.
find the second derivative, $g(x)$.
$g(x) =$

evaluate $g(-1)$.
$g(-1) =$

based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x = -1$?
at $x = -1$ the graph of $g(x)$ is select an answer

based on the concavity of $g(x)$ at $x = -1$, does this mean that there is a local minimum or local maximum at $x = -1$?
at $x = -1$ there is a local select an answer

Accepted Answer

{ "quality": "clear", "question_count": 5, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Derivative Definition", "Derivative Equations", "Concavity Analysis" ], "new_concepts": [ "Second Derivative Test" ], "current_concepts": [ "Derivative Definition", "Derivative Equations", "Concavity Analysis", "Second Derivative Test" ] } ## Find the first derivative $g'(x)$ $$ \begin{aligned} g(x) &= 6x^3 + 45x^2 + 72x \ g'(x) &= \frac{d}{dx}(6x^3 + 45x^2 + 72x) \ g'(x) &= 18x^2 + 90x + 72 \end{aligned} $$ ## Find the second derivative $g''(x)$ $$ \begin{aligned} g''(x) &= \frac{d}{dx}(18x^2 + 90x + 72) \ g''(x) &= 36x + 90 \end{aligned} $$ ## Evaluate $g''(-1)$ and determine concavity and local extrema $$ \begin{aligned} g''(-1) &= 36(-1) + 90 = 54 \ g''(-1) &= 54 > 0 \implies \text{concave up} \ g'(-1) = 0 \text{ and } g''(-1) > 0 &\implies \text{local minimum} \end{aligned} $$ | No. | Answer | |---|---| | 1 | $g'(x) = 18x^2 + 90x + 72$ | | 2 | $g''(x) = 36x + 90$ | | 3 | $g''(-1) = 54$ | | 4 | concave up | | 5 | local minimum | { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Calculus", "Second Derivative Test" ] }

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QUESTION IMAGE

Question

Explanation:

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

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

Evaluate \(g''(-1)\) and determine concavity and local extrema

Answer:

No.	Answer
2	\(g''(x) = 36x + 90\)
3	\(g''(-1) = 54\)
4	concave up
5	local minimum