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Consider the following function. Use a graphing utility to conflim your answers for parts (a) through (c). (If an answer does not exist, enter DNE.) \[ f(x)=\left\{\begin{array}{ll} 1 x+1, & x \leq-1 \\ x^{2}-4, & x>-1 \end{array}\right. \] (a) Find the critical numbers of \( f \). (Enter your answers as a comma-separated list.) \[ x=\square \] (b) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) increasing decreasing \( \square \) (c) Apply the first Derlvative Test to identify all relative extrema. relative maximum \( \quad(x, y)=( \) \( \square \) relative minimum \( \quad(x, y)=(\square) \) \( \square \)

Ask by Cross Mathis. in the United States
Mar 16,2025

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Answer

**(a) Critical Numbers:** \[ x = -1, 0 \] **(b) Intervals of Monotonicity:** - **Increasing:** \((-\infty, -1) \cup (0, \infty)\) - **Decreasing:** \((-1, 0)\) **(c) Relative Extrema:** - **Relative Maximum:** \((-1, 0)\) - **Relative Minimum:** \((0, -4)\)

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The Deep Dive

To get started, let's break down the piecewise function \( f(x) \) and evaluate its critical numbers, intervals of increase and decrease, and relative extrema. First off, look at the critical numbers. Critical numbers occur where the derivative is zero or undefined. 1. For \( x \leq -1 \): - The function is \( f(x) = x + 1 \). The derivative is \( f'(x) = 1 \), which is never zero or undefined. 2. For \( x > -1 \): - The function is \( f(x) = x^2 - 4 \). The derivative is \( f'(x) = 2x \). Setting \( 2x = 0 \) gives \( x = 0 \). Since \( x = 0 \) is in the domain \( x > -1\), it's a critical number. Additionally, check \( x = -1 \) since it’s where the pieces of the function meet. So, the critical numbers are \( -1, 0 \). --- Next, let's figure out the intervals of increase and decrease. - For \( x < -1 \): \( f'(x) = 1 \) (increasing). - For \( -1 < x < 0 \): \( f'(x) = 2x \): - Since \( f'(x) < 0 \) in this interval (e.g., at \( x = -0.5 \), \( f'(-0.5) = -1 \)), it's decreasing. - For \( x > 0 \): \( f'(x) = 2x \) (increasing). So, we find: - Increasing: \( (-\infty, -1) \cup (0, \infty) \) - Decreasing: \( (-1, 0) \) --- Finally, applying the First Derivative Test for relative extrema: - At \( x = -1 \): - Before \( -1 \) (increasing) to \( -1 \) (decreasing), we have a relative maximum: - \( f(-1) = 0 \), so the point is \( (-1, 0) \). - At \( x = 0 \): - Before \( 0 \) (decreasing) to \( 0 \) (increasing), we have a relative minimum: - \( f(0) = -4 \), so the point is \( (0, -4) \). So, the answers are: (a) \( x = -1, 0 \) (b) increasing \( (-\infty, -1) \cup (0, \infty) \), decreasing \( (-1, 0) \) (c) relative maximum \( (x, y) = (-1, 0) \), relative minimum \( (x, y) = (0, -4) \)

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