The function $g$ is defined and differentiable on the closed interval $[-6,6]$ and satisfies $g(0)=4$ . The graph of $y=g^{\prime}(x)$ , the derivative of $g$ , consists of a semicircle and three line segments, as shown in the figure below. $\newline$ (c) Find the $x$ -coordinate of each critical point of $g$ , where $-6<x<6$ , and classify each critical point as the location of a relative minimum, relative maximum, or neither a minimum nor a maximum. Explain your reasoning.

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Complex conjugate theorem

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Q. The function

g

is defined and differentiable on the closed interval

[-6,6]

and satisfies

g(0)=4

. The graph of

y=g^{\prime}(x)

, the derivative of

g

, consists of a semicircle and three line segments, as shown in the figure below.

\newline

x

-coordinate of each critical point of

g

, where

-6<x<6

, and classify each critical point as the location of a relative minimum, relative maximum, or neither a minimum nor a maximum. Explain your reasoning.

Identify Critical Points: Identify where the derivative $g'(x)$ equals $0$ or is undefined, as these are potential critical points. The graph of $g'(x)$ includes a semicircle and line segments.
Analyze Semicircle Part: Analyze the semicircle part of $g'(x)$ . If the semicircle is centered at the origin and spans from $-r$ to $r$ , then $g'(x) = 0$ at $x = 0$ , since this is the highest or lowest point of the semicircle depending on its orientation.
Check Endpoints of Segments: Check the endpoints of the line segments. If any segment is vertical, $g'(x)$ would be undefined at that $x$ -coordinate. If the graph is continuous and only consists of horizontal and non-vertical sloped lines, then $g'(x)$ does not equal zero at these points.
Determine Nature of Critical Point: Determine the nature of the critical point at $x = 0$ . If the semicircle is concave down (opening downwards), then $g'(x)$ changes from positive to negative as $x$ increases through $0$ , indicating a relative maximum. If concave up, it's a relative minimum.
Conclude Critical Point: Since no other critical points are indicated or suggested by the problem statement, and no vertical line segments are mentioned, conclude that $x = 0$ is the only critical point within the interval $(-6, 6)$ .