To solve for the values of the piecewise function [tex]\( f(x) \)[/tex] at the given points, we need to evaluate the function according to its definition for each specific value of [tex]\( x \)[/tex].
The piecewise function is defined as:
[tex]\[ f(x) = \begin{cases}
3x^2 & \text{if } x \leq -2 \\
2x & \text{if } x \geq 1
\end{cases} \][/tex]
Let's evaluate this function at [tex]\( x = 1 \)[/tex], [tex]\( x = -3 \)[/tex], and [tex]\( x = 0 \)[/tex] step-by-step.
### 1. Evaluating [tex]\( f(1) \)[/tex]:
We need to determine which part of the piecewise function applies to [tex]\( x = 1 \)[/tex]:
- Since [tex]\( 1 \geq 1 \)[/tex], we use the second case: [tex]\( f(x) = 2x \)[/tex].
Now, substituting [tex]\( x = 1 \)[/tex] into the function:
[tex]\[ f(1) = 2 \times 1 = 2 \][/tex]
### 2. Evaluating [tex]\( f(-3) \)[/tex]:
Next, we consider [tex]\( x = -3 \)[/tex]:
- Since [tex]\( -3 \leq -2 \)[/tex], we use the first case: [tex]\( f(x) = 3x^2 \)[/tex].
Now, substituting [tex]\( x = -3 \)[/tex] into the function:
[tex]\[ f(-3) = 3 \times (-3)^2 = 3 \times 9 = 27 \][/tex]
### 3. Evaluating [tex]\( f(0) \)[/tex]:
Finally, we evaluate [tex]\( x = 0 \)[/tex]:
- The value [tex]\( x = 0 \)[/tex] does not satisfy either [tex]\( x \leq -2 \)[/tex] or [tex]\( x \geq 1 \)[/tex]. Thus, [tex]\( f(x) \)[/tex] is not defined for [tex]\( x = 0 \)[/tex].
### Summary of Evaluations:
- [tex]\( f(1) = 2 \)[/tex]
- [tex]\( f(-3) = 27 \)[/tex]
- [tex]\( f(0) = \text{undefined} \)[/tex]
Therefore, the results are:
[tex]\[ f(1) = 2 \][/tex]
[tex]\[ f(-3) = 27 \][/tex]
[tex]\[ f(0) = \text{undefined} \][/tex]
