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Use the piecewise-defined function to find [tex]f(0)[/tex] and [tex]f(9)[/tex].

[tex]
f(x) = \left\{
\begin{array}{ll}
2 - 3x & \text{if } x \leq 3 \\
3x & \text{if } 3 \ \textless \ x \ \textless \ 8 \\
5x + 5 & \text{if } x \geq 8
\end{array}
\right.
[/tex]

[tex]f(0) = \square[/tex] (Simplify your answer.)

[tex]f(9) = \square[/tex] (Simplify your answer.)


Sagot :

To find [tex]\( f(0) \)[/tex] and [tex]\( f(9) \)[/tex] using the given piecewise-defined function:

[tex]\[ f(x) = \begin{cases} 2 - 3x & \text{if } x \leq 3 \\ 3x & \text{if } 3 < x < 8 \\ 5x + 5 & \text{if } x \geq 8 \end{cases} \][/tex]

Step-by-Step Solution:

1. Calculate [tex]\( f(0) \)[/tex]:

Determine which piece of the piecewise function [tex]\( x = 0 \)[/tex] falls under:
[tex]\[ x = 0 \quad (\text{since } 0 \leq 3) \][/tex]
Use the piece of the function [tex]\( f(x) = 2 - 3x \)[/tex]:
[tex]\[ f(0) = 2 - 3(0) = 2 \][/tex]
Therefore, [tex]\( f(0) = 2 \)[/tex].

2. Calculate [tex]\( f(9) \)[/tex]:

Determine which piece of the piecewise function [tex]\( x = 9 \)[/tex] falls under:
[tex]\[ x = 9 \quad (\text{since } 9 \geq 8) \][/tex]
Use the piece of the function [tex]\( f(x) = 5x + 5 \)[/tex]:
[tex]\[ f(9) = 5(9) + 5 = 45 + 5 = 50 \][/tex]
Therefore, [tex]\( f(9) = 50 \)[/tex].

In conclusion:
[tex]\[ f(0) = 2 \][/tex]
[tex]\[ f(9) = 50 \][/tex]