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The first 18 cupcakes cost [tex][tex]$\$[/tex]1.50[tex]$[/tex] each, and all cupcakes after that cost [tex]$[/tex]\[tex]$1$[/tex][/tex] each.

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

[tex]f(-3) = -27[/tex]


Sagot :

Let's determine [tex]\( f(-3) \)[/tex] for the piecewise function given by:

[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 are asked to find [tex]\( f(-3) \)[/tex].

1. First, identify which part of the piecewise function to use:
- Since [tex]\(-3\)[/tex] falls within the interval [tex]\([-3, 4)\)[/tex], we use [tex]\( f(x) = 2x^2 - 9 \)[/tex].

2. Substitute [tex]\( x = -3 \)[/tex] into the function [tex]\( 2x^2 - 9 \)[/tex]:

[tex]\[ f(-3) = 2(-3)^2 - 9 \][/tex]

3. Calculate the value step-by-step:
- Calculate [tex]\((-3)^2\)[/tex]:

[tex]\[ (-3)^2 = 9 \][/tex]

- Multiply by 2:

[tex]\[ 2 \cdot 9 = 18 \][/tex]

- Subtract 9:

[tex]\[ 18 - 9 = 9 \][/tex]

Thus, the value of [tex]\( f(-3) \)[/tex] is [tex]\( 9 \)[/tex].

So, [tex]\( f(-3) = 9 \)[/tex].