Question

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:

To solve this, we need the original function \( g(x) \) (or \( g'(x) \)) which is not provided in the question. Without the original function, we cannot compute the second derivative \( g''(x) \), evaluate \( g''(3) \), or determine concavity and local extrema.

Step 1: Identify the missing information

We need the expression for \( g(x) \) (or \( g'(x) \)) to compute the second derivative. For example, if \( g(x) = x^3 - 9x^2 + 24x \), then:

Step 1: Find \( g'(x) \)

\( g'(x) = 3x^2 - 18x + 24 \)

Step 2: Find \( g''(x) \)

Differentiate \( g'(x) \): \( g''(x) = 6x - 18 \)

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

Substitute \( x = 3 \): \( g''(3) = 6(3) - 18 = 18 - 18 = 0 \) (but this is just an example; the actual value depends on \( g(x) \))

Step 4: Determine concavity

If \( g''(3) > 0 \), concave up (local minimum); if \( g''(3) < 0 \), concave down (local maximum); if \( g''(3) = 0 \), the test is inconclusive.

Answer:

To solve, please provide the original function \( g(x) \) (or \( g'(x) \)) to compute the second derivative, evaluate \( g''(3) \), and determine concavity and local extrema.