Question

given the function \\(g(x) = 6x^3 - 27x^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

Explanation:

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

$$ g'(x) = \frac{d}{dx}(6x^3 - 27x^2 - 72x) = 18x^2 - 54x - 72 $$

Find the second derivative \(g''(x)\) and evaluate \(g''(-1)\)

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Determine concavity and classify the local extremum at \(x = -1\)

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Answer:

Question 1

\(g'(x) = 18x^2 - 54x - 72\)

Question 2

\(g''(x) = 36x - 54\)

Question 3

\(g''(-1) = -90\)

Question 4

At \(x = -1\) the graph of \(g(x)\) is concave down

Question 5

At \(x = -1\) there is a local maximum