Math Problem Statement
Evaluate the function f(x) at each specified value of the independent variable and simplify: (a) f(-4), (b) f(0), (c) f(1).
Solution
The function is piecewise-defined as follows:
\begin{cases} 9 - 2x, & x < 0 \\ 9, & 0 \leq x < 1 \\ 8x + 1, & x \geq 1 \end{cases}$$ We need to evaluate $$f(x)$$ at the following values: ### (a) $$f(-4)$$ Since $$-4 < 0$$, we use the first case: $$f(x) = 9 - 2x$$. $$f(-4) = 9 - 2(-4) = 9 + 8 = 17$$ ### (b) $$f(0)$$ Since $$0 \leq x < 1$$, we use the second case: $$f(x) = 9$$. $$f(0) = 9$$ ### (c) $$f(1)$$ Since $$x = 1 \geq 1$$, we use the third case: $$f(x) = 8x + 1$$. $$f(1) = 8(1) + 1 = 8 + 1 = 9$$ ### Final Answers: - (a) $$f(-4) = 17$$ - (b) $$f(0) = 9$$ - (c) $$f(1) = 9$$ Would you like a detailed explanation of any step? Here are some related questions for practice: 1. What would $$f(x)$$ be for $$x = 2$$? 2. How would the value change for $$f(-1)$$? 3. What happens if $$x = 0.5$$? 4. Can you define $$f(x)$$ for all real numbers based on this function? 5. How does the slope of the function change in each piece? **Tip:** Always check which part of the piecewise function applies based on the input range.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Function Evaluation
Formulas
f(x) = 9 - 2x for x < 0
f(x) = 9 for 0 ≤ x < 1
f(x) = 8x + 1 for x ≥ 1
Theorems
Definition of Piecewise Functions
Suitable Grade Level
Grades 10-12