https://file.aiquickdraw.com/mathbot/file/17291478653067m3766su.jpg

Let's work through the problem presented in the image. The function \( f(x) \) has two parts:

\[
f(x) = \begin{cases} 
-x^2 + 1 & \text{for } x \leq 1 \\
2x & \text{for } x > 1 
\end{cases}
\]

### a. Graph the function
The function consists of two pieces:
1. \( f(x) = -x^2 + 1 \) for \( x \leq 1 \): This is a downward-opening parabola.
2. \( f(x) = 2x \) for \( x > 1 \): This is a straight line with a slope of 2.

The graph will consist of:
- A parabola segment from the left, stopping at \( x = 1 \).
- A straight line starting from \( x = 1 \) and extending to the right.
  
I can't draw graphs directly here, but you can sketch this by noting:
- For \( f(x) = -x^2 + 1 \), the vertex is at \( (0, 1) \), and it crosses the x-axis at \( x = -1 \).
- The line \( f(x) = 2x \) will pass through the point \( (1, 2) \) and will rise linearly afterward.

### b. Domain in interval notation
The domain is the set of all possible \( x \)-values. Since there is no restriction on \( x \), the domain is:

\[
\text{Domain: } (-\infty, \infty)
\]

### c. Range in interval notation
For the range:
- For \( x \leq 1 \), the parabola \( f(x) = -x^2 + 1 \) has a maximum value of 1 at \( x = 0 \) and decreases as \( x \) moves away from zero.
- For \( x > 1 \), the linear function \( f(x) = 2x \) starts at 2 and increases without bound.

Thus, the range is:

\[
\text{Range: } (-\infty, 1] \cup [2, \infty)
\]

### d. Evaluate \( f(-1) \), \( f(1) \), and \( f(2) \)
- \( f(-1) \): Use \( f(x) = -x^2 + 1 \) since \( x = -1 \leq 1 \).
  \[
  f(-1) = -(-1)^2 + 1 = -1 + 1 = 0
  \]

- \( f(1) \): Again, use \( f(x) = -x^2 + 1 \) since \( x = 1 \leq 1 \).
  \[
  f(1) = -(1)^2 + 1 = -1 + 1 = 0
  \]

- \( f(2) \): Use \( f(x) = 2x \) since \( x = 2 > 1 \).
  \[
  f(2) = 2 \times 2 = 4
  \]

### e. Solve \( f(x) = 6 \)
For \( f(x) = 6 \):
- If \( x > 1 \), use \( f(x) = 2x \):
  \[
  2x = 6 \quad \Rightarrow \quad x = 3
  \]
There is no solution for \( x \leq 1 \) because \( f(x) = -x^2 + 1 \leq 1 \), and \( 6 \) is outside this range.

Thus, \( x = 3 \).

### f. Solve \( f(x) = -3 \)
For \( f(x) = -3 \):
- Use \( f(x) = -x^2 + 1 \) for \( x \leq 1 \):
  \[
  -x^2 + 1 = -3 \quad \Rightarrow \quad -x^2 = -4 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = -2
  \]

Thus, \( x = -2 \).

### g. Intervals of increasing, decreasing, or constant behavior
- For \( x \leq 1 \), \( f(x) = -x^2 + 1 \) is a downward-opening parabola, so it decreases as \( x \) increases.
- For \( x > 1 \), \( f(x) = 2x \) is an increasing linear function.

So, the function is:
- **Decreasing** on \( (-\infty, 1] \).
- **Increasing** on \( (1, \infty) \).

### Do you have any questions or would you like further details?

Here are five related questions to deepen your understanding:
1. How does the vertex of a parabola affect its graph?
2. Why can't \( f(x) = -x^2 + 1 \) ever reach 6?
3. How would you determine if a function is continuous at \( x = 1 \)?
4. What happens to the graph of a quadratic function as the coefficient of \( x^2 \) changes?
5. How would the graph change if the linear part of the function had a different slope?

**Tip:** When solving piecewise functions, always check which part of the function applies to a given input value to avoid mistakes.

Graph the function and answer various questions about the piecewise-defined functions f(x) = {-x^2 + 1 for x ≤ 1, 2x for x > 1} and f(x) = {|x| for x < 2, -x for x ≥ 2}.

Learn how to graph and analyze piecewise functions such as f(x) = -x^2 + 1 and f(x) = 2x. This guide explores domain, range, and behavior across different intervals, including quadratic, linear, and absolute value functions. Suitable for high school students in Grades 9-12.

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