Answer :
To find [tex]\( f(5) \)[/tex] for the piecewise function
[tex]\[ f(x) = \begin{cases} x^3, & \text{if } x < -3 \\ 2x^2 - 9, & \text{if } -3 \leq x < 4 \\ 5x + 4, & \text{if } x \geq 4 \end{cases} \][/tex]
we need to determine which piece of the piecewise function to use based on the value of [tex]\( x \)[/tex].
Given [tex]\( x = 5 \)[/tex]:
1. We check the conditions for [tex]\( x \)[/tex]:
- [tex]\( x < -3 \)[/tex]: This is not true for [tex]\( x = 5 \)[/tex].
- [tex]\(-3 \leq x < 4\)[/tex]: This is not true for [tex]\( x = 5 \)[/tex].
- [tex]\( x \geq 4 \)[/tex]: This condition is true for [tex]\( x = 5 \)[/tex].
Since the condition [tex]\( x \geq 4 \)[/tex] is satisfied, we use the piece of the function [tex]\( f(x) = 5x + 4 \)[/tex].
2. Substitute [tex]\( x = 5 \)[/tex] into the function [tex]\( 5x + 4 \)[/tex]:
[tex]\[ f(5) = 5(5) + 4 \][/tex]
3. Perform the arithmetic:
[tex]\[ f(5) = 25 + 4 = 29 \][/tex]
Thus, the value of [tex]\( f(5) \)[/tex] is [tex]\( 29 \)[/tex].
[tex]\[ f(x) = \begin{cases} x^3, & \text{if } x < -3 \\ 2x^2 - 9, & \text{if } -3 \leq x < 4 \\ 5x + 4, & \text{if } x \geq 4 \end{cases} \][/tex]
we need to determine which piece of the piecewise function to use based on the value of [tex]\( x \)[/tex].
Given [tex]\( x = 5 \)[/tex]:
1. We check the conditions for [tex]\( x \)[/tex]:
- [tex]\( x < -3 \)[/tex]: This is not true for [tex]\( x = 5 \)[/tex].
- [tex]\(-3 \leq x < 4\)[/tex]: This is not true for [tex]\( x = 5 \)[/tex].
- [tex]\( x \geq 4 \)[/tex]: This condition is true for [tex]\( x = 5 \)[/tex].
Since the condition [tex]\( x \geq 4 \)[/tex] is satisfied, we use the piece of the function [tex]\( f(x) = 5x + 4 \)[/tex].
2. Substitute [tex]\( x = 5 \)[/tex] into the function [tex]\( 5x + 4 \)[/tex]:
[tex]\[ f(5) = 5(5) + 4 \][/tex]
3. Perform the arithmetic:
[tex]\[ f(5) = 25 + 4 = 29 \][/tex]
Thus, the value of [tex]\( f(5) \)[/tex] is [tex]\( 29 \)[/tex].