Question

Math

Posted 3 months ago

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The twice differentiable function $g$ and its second derivative $g^{\prime \prime}$ are graphed.
Given that $g^{\prime}(c)=0$, what is an appropriate calculus-based justification for the fact that $g$ has a relative maximum at $x=c$ ?
Choose 1 answer:
(A) $g^{\prime \prime}(c)<0$
(B) $g$ is concave down at $x=c$.
(c) $g^{\prime \prime}$ is concave up at $x=c$.
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 2

Since $g^{\prime}(c) = 0$, $x = c$ is a critical point. To determine if it is a relative maximum, we examine the second derivative $g^{\prime \prime}(x)$

step 3

From the graph, $g^{\prime \prime}(x)$ has a zero at $x = c$ and changes sign from negative to positive as $x$ increases through $c$

step 4

This sign change indicates that $g(x)$ is concave down before $x = c$ and concave up after $x = c$

step 5

Therefore, $g(x)$ has a relative maximum at $x = c$ because $g^{\prime \prime}(c) < 0$ and $g$ is concave down at $x = c$

B

Key Concept

Concavity and Second Derivative Test

Explanation

The second derivative test states that if $g^{\prime}(c) = 0$ and $g^{\prime \prime}(c) < 0$, then $g$ has a relative maximum at $x = c$ because the function is concave down at that point.

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