Math Problem Statement
Evaluate the function for the given values of x, where f(x) is defined piecewise as follows:
-3x + 6 for x ≤ -1
x^2 + 3 for -1 ≤ x < 5
6 for x ≥ 5
Solution
It looks like you've uploaded an image of a piecewise function. The function is defined as:
\begin{cases} -3x + 6 & \text{for } x \leq -1 \\ x^2 + 3 & \text{for } -1 \leq x < 5 \\ 6 & \text{for } x \geq 5 \end{cases}$$ Now, let's evaluate the function for the given values of $$x$$: 1. **Part 1**: $$f(4)$$ - Since $$4$$ is within the range $$-1 \leq x < 5$$, we use the second case, $$f(x) = x^2 + 3$$. $$f(4) = 4^2 + 3 = 16 + 3 = 19$$ 2. **Part 2**: $$f(-2)$$ - Since $$-2$$ is within the range $$-1 \leq x < 5$$, we again use the second case, $$f(x) = x^2 + 3$$. $$f(-2) = (-2)^2 + 3 = 4 + 3 = 7$$ 3. **Part 3**: $$f(-1)$$ - Since $$-1$$ is within the range $$-1 \leq x < 5$$, we use the second case again. $$f(-1) = (-1)^2 + 3 = 1 + 3 = 4$$ Let me know if you'd like any further details or explanations on these steps! Here are some related questions you may find useful: 1. How to evaluate a piecewise function at different values of $$x$$? 2. How do the values of $$x$$ determine which piece of the function to use? 3. What happens when $$x$$ is equal to a boundary value, like $$x = -1$$ or $$x = 5$$? 4. How would the function behave for values outside the given ranges? 5. Can you explain why the second part of the piecewise function uses $$x^2 + 3$$? **Tip**: Always identify the correct range for $$x$$ first when working with piecewise functions, then use the corresponding expression to evaluate the function.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Polynomial Functions
Algebra
Formulas
f(x) = -3x + 6
f(x) = x^2 + 3
f(x) = 6
Theorems
Piecewise Function Evaluation
Suitable Grade Level
Grades 8-10