relative maximum justification using first derivative test
Check the final answer first, then review the worked steps.
Answer
By the First Derivative Test, $g$ has a relative maximum at $x = c$ because $g'(c) = 0$ and $g'(x)$ changes from positive to negative at $x = c$.
Step-by-step solution
- Identify the critical point: A relative maximum or minimum of a differentiable function $g$ occurs at a critical point where the derivative $g'(x) = 0$. Looking at the graph, the brown line representing $g'$ crosses the x-axis at $x = c$, meaning $g'(c) = 0$.
- Analyze the sign change of the derivative: According to the First Derivative Test, if $g'(x)$ changes from positive to negative at a critical point $x = c$, then $g$ has a relative maximum at $x = c$.
- Observe the graph of $g'$: To the left of $x = c$ (where $x < c$), the graph of $g'$ is above the x-axis, meaning $g'(x) > 0$. To the right of $x = c$ (where $x > c$), the graph of $g'$ is below the x-axis, meaning $g'(x) < 0$.
- Formulate the justification: Since $g'(c) = 0$ and $g'(x)$ changes sign from positive to negative at $x = c$, the function $g$ must have a relative maximum at $x = c$ by the First Derivative Test.