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 $$ \quad(x, y)=(\square) $$ $$ \square $$

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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 $$ \quad(x, y)=(\square) $$ $$ \square $$

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**(a) Finding the Critical Numbers** A critical number is a number in the domain where either the derivative is zero or does not exist. For $$ f(x)=\begin{cases} x+1,& x\le -1,\ x^2-4,& x>-1, \end{cases} $$ we consider both pieces. 1. For $$x\le -1$$: $$ f(x)=x+1 \quad\text{with} \quad f'(x)=1. $$ Since $$f'(x)=1$$ is never zero and is defined for all $$x\le-1$$, there are no critical numbers from this piece, except we must check the boundary. 2. For $$x>-1$$: $$ f(x)=x^2-4 \quad\text{with} \quad f'(x)=2x. $$ Set the derivative equal to zero: $$ 2x=0\quad\Rightarrow\quad x=0. $$ Since $$0 > -1$$, $$x=0$$ is a critical number. 3. At the boundary $$x=-1$$: - From the left: $$f'(x)=1$$. - From the right: $$f'(x)=2(-1)=-2$$. Because the left-hand derivative and right-hand derivative do not agree, $$f'(x)$$ does not exist at $$x=-1$$. Thus, $$x=-1$$ is also a critical number. So the critical numbers are: $$ x=-1,\;0. $$ --- **(b) Finding Intervals of Increase and Decrease** We now determine the sign of the derivative on each interval created by the critical numbers. 1. **For $$x<-1$$:** Here, $$f(x)=x+1$$ and $$f'(x)=1$$. Since $$1>0$$, the function is increasing on this interval. $$\textbf{Increasing on: }(-\infty,-1)$$. 2. **For $$-10$$:** Here, $$f'(x)=2x$$ is positive (since $$x>0$$), hence the function is increasing. $$\textbf{Increasing on: }(0,\infty)$$. Thus, in interval notation: - Increasing on $$ (-\infty,-1) \cup (0,\infty) $$. - Decreasing on $$ (-1,0) $$. --- **(c) Identifying Relative Extrema Using the First Derivative Test** Examine changes in the sign of the derivative at the critical points: 1. **At $$x=-1$$:** - For $$x<-1$$, $$f'(x)=1 > 0$$ (increasing). - For $$-10$$, $$f'(x)=2x > 0$$ (increasing). Since $$f$$ goes from decreasing to increasing at $$x=0$$, there is a relative minimum there. The value is: $$ f(0)=0^2-4=-4. $$ So, the relative minimum is $$(0,-4)$$. --- **Final Answers** (a) Critical numbers: $$ x=-1,\;0 $$ (b) Intervals of monotonicity: - Increasing: $$(-\infty,-1) \cup (0,\infty)$$ - Decreasing: $$(-1,0)$$ (c) Relative extrema: - Relative maximum at $$(-1,0)$$ - Relative minimum at $$(0,-4)$$ **(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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