Question

use technology to graph the function in order to answer the question below.
$f(x)=\

$$\begin{cases}-x-3, & x < -1 \\\\ \\sqrt{x+5}-4, & -1 \\leq x < 4 \\\\ (x-4)^2-1, & 4 \\leq x \\leq 5\\end{cases}$$

$
using the list of intervals below, choose the interval(s) to where the function $f$ is:
positive
increasing
negative
decreasing
$(-3,5)$
$(-\infty, -1)$
$(4,5)$
$(-1,4)\cup(4,5)$
$(-\infty, -3)$

Explanation:

Step1: Analyze positive intervals

For $f(x)>0$:

1. For $x<-1$: $-x-3>0 \implies x<-3$
2. For $-1\leq x<4$: $\sqrt{x+5}-4>0 \implies \sqrt{x+5}>4 \implies x+5>16 \implies x>11$, which has no overlap with $-1\leq x<4$
3. For $4\leq x\leq5$: $(x-4)^2-1>0 \implies (x-4)^2>1 \implies x-4>1 \text{ or } x-4<-1$. With $4\leq x\leq5$, this gives $x>5$ (no overlap) or $x<3$ (no overlap).

Only interval: $(-\infty,-3)$

Step2: Analyze increasing intervals

Check slopes of each piece:

1. For $x<-1$: $f(x)=-x-3$, slope $-1$ (decreasing)
2. For $-1\leq x<4$: $f(x)=\sqrt{x+5}-4$, derivative $\frac{1}{2\sqrt{x+5}}>0$ (increasing)
3. For $4\leq x\leq5$: $f(x)=(x-4)^2-1$, derivative $2(x-4)\geq0$ (increasing for $x\geq4$)

Combined interval: $(-1,4)\cup(4,5)$

Step3: Analyze negative intervals

For $f(x)<0$:

1. For $x<-1$: $-x-3<0 \implies x>-3$, so $-3
2. For $-1\leq x<4$: $\sqrt{x+5}-4<0 \implies \sqrt{x+5}<4 \implies x+5<16 \implies x<11$, which holds for all $-1\leq x<4$
3. For $4\leq x\leq5$: $(x-4)^2-1<0 \implies (x-4)^2<1 \implies -1

Combined interval: $(-3,5)$

Step4: Analyze decreasing intervals

Check slopes of each piece:

1. For $x<-1$: $f(x)=-x-3$, slope $-1$ (decreasing)
2. For $-1\leq x<4$: increasing (from Step2)
3. For $4\leq x\leq5$: increasing (from Step2)

Only interval: $(-\infty,-1)$

Answer:

Positive: $(-\infty,-3)$
Increasing: $(-1,4)\cup(4,5)$
Negative: $(-3,5)$
Decreasing: $(-\infty,-1)$