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)

Q: 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 $x=c$.

∻Solution by Steps∻ ‖ step 1 ⋮ Given that $g$ is a twice differentiable function and $g^{\prime}(c) = 0$, we need to determine the nature of the critical point at $x = c$. ‖ ‖ 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)

Question

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 $x=c$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given that $g$ is a twice differentiable function and $g^{\prime}(c) = 0$, we need to determine the nature of the critical point at $x = c$. ‖ ‖ 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.⚹◊What is the relationship between the concavity of a function $g$ and the sign of its second derivative $g''$ at a point $x=c$ where the first derivative is zero, $g'(c)=0$?⍭ Generate me a similar question◊

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