Use a graph to determine the local minimum(s) of \( g(x)=\left\{\begin{array}{ll}|x| & \text { if }-2 \leq x \leq 1 \\ \sqrt{x} & \text { if } 1

Ask by Lyons Simpson. in the United States

Jan 22,2025

To determine the local minima of the piecewise function \[ g(x) = \begin{cases} |x| & \text{if } -2 \leq x \leq 1, \\ \sqrt{x} & \text{if } 1 < x < 4, \\ -x + 6 & \text{if } 4 \leq x \leq 8, \end{cases} \] let's analyze each interval and the behavior of the function at the transition points. ### 1. Interval \([-2, 1]\): \(g(x) = |x|\) - **Graph Shape**: V-shaped with the vertex at \(x = 0\). - **Local Minimum**: At \(x = 0\), \(g(0) = 0\). This is the lowest point in this interval. ### 2. Interval \(1 < x < 4\): \(g(x) = \sqrt{x}\) - **Graph Shape**: Increasing function. - **Behavior at \(x = 1\)**: \(g(1) = 1\) (continuous with the previous interval). - **Derivative from Left (\(x < 1\))**: \(1\). - **Derivative from Right (\(x > 1\))**: \(\frac{1}{2\sqrt{x}} = \frac{1}{2}\). - **Conclusion**: \(x = 1\) is not a local minimum since the function continues to increase on both sides. ### 3. Interval \(4 \leq x \leq 8\): \(g(x) = -x + 6\) - **Graph Shape**: Decreasing linear function. - **Behavior at \(x = 4\)**: \(g(4) = 2\). - **Derivative from Left (\(x < 4\))**: \(\frac{1}{4}\). - **Derivative from Right (\(x > 4\))**: \(-1\). - **Conclusion**: \(x = 4\) is a local maximum since the function changes from increasing to decreasing. ### 4. Endpoint \(x = 8\) - **Value**: \(g(8) = -8 + 6 = -2\). - **Behavior**: As \(x\) approaches 8 from the left, the function decreases towards \(-2\). - **Conclusion**: \(x = 8\) is a local minimum because it is the lowest point in its immediate neighborhood. ### Summary of Local Minima: - **\((0, 0)\)**: Local minimum from the first interval. - **\((8, -2)\)**: Local minimum at the endpoint. Therefore, the only local minima of \(g(x)\) are at \((0, 0)\) and \((8, -2)\). **Answer:** \( (0,0) \) and \( (8,-2) \) are the only local minimums