Math Problem Statement
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}.
Solution
Let's work through the problem presented in the image. The function has two parts:
-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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Quadratic Functions
Linear Functions
Absolute Value Functions
Domain and Range
Intervals of Increase and Decrease
Formulas
f(x) = -x^2 + 1
f(x) = 2x
f(x) = |x|
f(x) = -x
Theorems
Piecewise Function Behavior
Graphing Quadratic and Linear Functions
Suitable Grade Level
Grades 9-12