To solve this problem, we need to evaluate the given piecewise function [tex]\( f(x) \)[/tex] at various points within the specified ranges and then check which of the values listed are true.
The piecewise function is given by:
[tex]\[ f(x) = \begin{cases}
x^2 & \text{if } x \leq 3 \\
2x - 4 & \text{if } x > 3
\end{cases} \][/tex]
Let's go through each of the options one by one:
a. [tex]\( f(3) = 9 \)[/tex]
Since [tex]\( 3 \leq 3 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(3) = 3^2 = 9 \][/tex]
Hence, this is true.
b. [tex]\( f(0) = -4 \)[/tex]
Since [tex]\( 0 \leq 3 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(0) = 0^2 = 0 \][/tex]
Hence, this is false.
c. [tex]\( f(0) = 0 \)[/tex]
Since [tex]\( 0 \leq 3 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(0) = 0^2 = 0 \][/tex]
Hence, this is true.
d. [tex]\( f(4) = 16 \)[/tex]
Since [tex]\( 4 > 3 \)[/tex], we use the second part of the piecewise function:
[tex]\[ f(4) = 2 \cdot 4 - 4 = 8 - 4 = 4 \][/tex]
Hence, this is false.
e. [tex]\( f(3) = 2 \)[/tex]
As previously calculated for option (a), since [tex]\( 3 \leq 3 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(3) = 3^2 = 9 \][/tex]
Hence, this is false.
f. [tex]\( f(-1) = 1 \)[/tex]
Since [tex]\( -1 \leq 3 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(-1) = (-1)^2 = 1 \][/tex]
Hence, this is true.
So, the true values among the given choices are:
- a. [tex]\( f(3) = 9 \)[/tex]
- c. [tex]\( f(0) = 0 \)[/tex]
- f. [tex]\( f(-1) = 1 \)[/tex]
