Given the piecewise function below, choose the true values.

[tex]
f(x) = \left\{\begin{array}{ll}
x^2 & \text{if } x \leq 3 \\
2x - 4 & \text{if } x \ \textgreater \ 3
\end{array}\right.
[/tex]

Select one or more:

a. [tex]f(3)=9[/tex]

b. [tex]f(0)=-4[/tex]

c. [tex]f(0)=0[/tex]

d. [tex]f(4)=16[/tex]

e. [tex]f(3)=2[/tex]

f. [tex]f(-1)=1[/tex]



Answer :

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]