Determine [tex]\( f(3) \)[/tex] for [tex]\( f(x)=\left\{
\begin{array}{cc}
x^3, & x\ \textless \ -4 \\
2x^2 - 9, & -4 \leq x \ \textless \ 3 \\
5x + 4, & x \geq 3
\end{array}
\right. \)[/tex]

A. 9

B. 19

C. 22

D. 27



Answer :

To determine [tex]\( f(3) \)[/tex] for the given piecewise function [tex]\( f(x) \)[/tex], let's carefully consider which part of the function applies to [tex]\( x = 3 \)[/tex]:

The function is defined as:
[tex]\[ f(x) = \begin{cases} x^3, & \text{if } x < -4 \\ 2x^2 - 9, & \text{if } -4 \leq x < 3 \\ 5x + 4, & \text{if } x \geq 3 \end{cases} \][/tex]

Now, we need to find [tex]\( f(3) \)[/tex]:

1. First case: [tex]\( x^3 \)[/tex] for [tex]\( x < -4 \)[/tex]
- This does not apply as [tex]\( x = 3 \)[/tex] is not less than [tex]\(-4\)[/tex].

2. Second case: [tex]\( 2x^2 - 9 \)[/tex] for [tex]\( -4 \leq x < 3 \)[/tex]
- This does not apply as [tex]\( x = 3 \)[/tex] is not less than 3 (it is actually equal to 3).

3. Third case: [tex]\( 5x + 4 \)[/tex] for [tex]\( x \geq 3 \)[/tex]
- This applies because [tex]\( x = 3 \)[/tex] is greater than or equal to 3.

Therefore, we will use the third case [tex]\( 5x + 4 \)[/tex]:
[tex]\[ f(3) = 5(3) + 4 = 15 + 4 = 19 \][/tex]

Thus, the value of [tex]\( f(3) \)[/tex] is [tex]\( 19 \)[/tex].

[tex]\(\boxed{19}\)[/tex]