Use Desmos to graph the function 1. f(x)= -x-3,&x<-1\ sqrt x+5 -4,&-1<= x<4\ (x - 4) ^ 2 - 1 ,&4<= x<=5 the function the following key features. Use a. For what interval(s) is the function increasing? b. For what interval(s) is the function decreasing? c. For what interval(s) is the function positive? d. For what interval(s) is the function negative?

The function you've provided is a piecewise function defined as follows:

\[
f(x) = \begin{cases} 
-x - 3 & \text{for } x < -1 \\
\sqrt{x + 5} - 4 & \text{for } -1 \leq x < 4 \\
(x - 4)^2 - 1 & \text{for } 4 \leq x \leq 5
\end{cases}
\]

To answer the questions, let's break down the problem:

### a. Interval(s) where the function is increasing:

- **First piece:** \(f(x) = -x - 3\) for \(x < -1\):
  - The slope is \(-1\), so this part is decreasing.
  
- **Second piece:** \(f(x) = \sqrt{x + 5} - 4\) for \(-1 \leq x < 4\):
  - The derivative is \(f'(x) = \frac{1}{2\sqrt{x + 5}}\), which is positive for \(x \geq -1\), so this part is increasing.

- **Third piece:** \(f(x) = (x - 4)^2 - 1\) for \(4 \leq x \leq 5\):
  - The derivative is \(f'(x) = 2(x - 4)\), which is positive for \(x > 4\), so this part is increasing for \(4 < x \leq 5\).

**The function is increasing on the intervals \([-1, 4)\) and \((4, 5]\).**

### b. Interval(s) where the function is decreasing:

- **First piece:** \(f(x) = -x - 3\) for \(x < -1\):
  - As mentioned, the slope is \(-1\), so this part is decreasing.

- **Second piece:** \(f(x) = \sqrt{x + 5} - 4\) for \(-1 \leq x < 4\):
  - This part is increasing, not decreasing.

- **Third piece:** \(f(x) = (x - 4)^2 - 1\) for \(4 \leq x \leq 5\):
  - The derivative is \(f'(x) = 2(x - 4)\), which is negative for \(x < 4\), but since the interval starts at \(x = 4\), there's no decreasing interval in this part.

**The function is decreasing on the interval \((-\infty, -1)\).**

### c. Interval(s) where the function is positive:

- **First piece:** \(f(x) = -x - 3\) for \(x < -1\):
  - This part is negative for all \(x\).

- **Second piece:** \(f(x) = \sqrt{x + 5} - 4\) for \(-1 \leq x < 4\):
  - Set \(f(x) > 0\): \(\sqrt{x + 5} - 4 > 0 \Rightarrow \sqrt{x + 5} > 4 \Rightarrow x + 5 > 16 \Rightarrow x > 11\), but since \(x\) is in \([-1, 4)\), this part does not contribute to positive values.

- **Third piece:** \(f(x) = (x - 4)^2 - 1\) for \(4 \leq x \leq 5\):
  - Set \(f(x) > 0\): \((x - 4)^2 > 1\), so \(x - 4 > 1\) or \(x - 4 < -1\), but within \([4, 5]\), only \(x > 5\) works which is outside the interval, hence this part doesn't contribute either.

**The function does not have any intervals where it is positive.**

### d. Interval(s) where the function is negative:

- **First piece:** \(f(x) = -x - 3\) for \(x < -1\):
  - Always negative as \(x\) is negative and slope is negative.

- **Second piece:** \(f(x) = \sqrt{x + 5} - 4\) for \(-1 \leq x < 4\):
  - Set \(f(x) < 0\): \(\sqrt{x + 5} < 4 \Rightarrow x + 5 < 16 \Rightarrow x < 11\), but since \(x\) is in \([-1, 4)\), this part does contribute to negative values for all \(x < 11\).

- **Third piece:** \(f(x) = (x - 4)^2 - 1\) for \(4 \leq x \leq 5\):
  - Set \(f(x) < 0\): \((x - 4)^2 < 1\) leads to \(-1 < x - 4 < 1\), so \(3 < x < 5\), but since the interval is \([4, 5]\), it covers the region where the function is negative at \(4\) and increasing to \(0\) at \(5\).

**The function is negative on the intervals \((-\infty, -1)\), \([-1, 4)\), and at \(4\).**

Would you like further details or have any other questions? Here are some related questions you might find interesting:

1. How do you find the derivative of a piecewise function?
2. How can you determine the concavity of each piece of the function?
3. What is the significance of the function's behavior at the boundary points?
4. How would the graph of this function look like?
5. What if we extended the function's definition to a larger domain?

**Tip:** When working with piecewise functions, always check the continuity and differentiability at the boundary points.

Explore how to analyze the behavior of a piecewise function including intervals where it increases, decreases, is positive, and negative. Understand the derivative formulas for each piece and their implications. Suitable for advanced high school students and beyond.

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