Math Problem Statement
Suppose the function h(x) is defined as a piecewise function with three cases. Find h(-2), h(0), and h(3).
Solution
We are tasked with evaluating the piecewise-defined function at three points: , , and . The function is defined as follows:
\begin{cases} \frac{1}{4}x - 1 & \text{if } x < -2, \\ (x+1)^2 - 1 & \text{if } -2 \leq x < 2, \\ -\frac{1}{2}x + 2 & \text{if } x \geq 2. \end{cases}$$ ### Step-by-step evaluation: 1. **Evaluate $$h(-2)$$:** - Since $$-2$$ falls into the interval $$-2 \leq x < 2$$, we use the second case of the piecewise function: $$h(x) = (x+1)^2 - 1$$. - Substituting $$x = -2$$ into this expression: $$h(-2) = (-2+1)^2 - 1 = (-1)^2 - 1 = 1 - 1 = 0.$$ Therefore, $$h(-2) = 0$$. 2. **Evaluate $$h(0)$$:** - Since $$0$$ also falls into the interval $$-2 \leq x < 2$$, we use the same expression as before: $$h(x) = (x+1)^2 - 1$$. - Substituting $$x = 0$$: $$h(0) = (0+1)^2 - 1 = 1^2 - 1 = 1 - 1 = 0.$$ Therefore, $$h(0) = 0$$. 3. **Evaluate $$h(3)$$:** - Since $$3 \geq 2$$, we use the third case of the function: $$h(x) = -\frac{1}{2}x + 2$$. - Substituting $$x = 3$$: $$h(3) = -\frac{1}{2}(3) + 2 = -\frac{3}{2} + 2 = -\frac{3}{2} + \frac{4}{2} = \frac{1}{2}.$$ Therefore, $$h(3) = \frac{1}{2}$$. ### Final answers: - $$h(-2) = 0$$, - $$h(0) = 0$$, - $$h(3) = \frac{1}{2}$$. Would you like any further explanation or additional details on this? --- Here are 5 related questions to explore: 1. How do you determine which piece of a piecewise function to use for a given $$x$$? 2. What happens if a piecewise function has more than three cases? 3. How do you graph a piecewise-defined function? 4. Can a piecewise function be discontinuous? If so, how can you tell? 5. What are the applications of piecewise functions in real-life problems? **Tip:** Always check the intervals of the piecewise function carefully to ensure you are using the correct expression for the given input.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Function Evaluation
Formulas
h(x) = (1/4)x - 1, if x < -2
h(x) = (x+1)^2 - 1, if -2 ≤ x < 2
h(x) = -(1/2)x + 2, if x ≥ 2
Theorems
Piecewise Function Evaluation
Suitable Grade Level
Grades 9-11