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The equation of the piecewise defined function [tex]f(x)[/tex] is below. What is the value of [tex]f(1)[/tex]?

[tex]\[ f(x) = \left\{
\begin{array}{cl}
x^2 + 1, & -4 \leq x \ \textless \ 1 \\
-x^2, & 1 \leq x \ \textless \ 2 \\
3x, & x \geq 2
\end{array}
\right. \][/tex]

A. [tex]f(1) = -2[/tex]
B. [tex]f(1) = -1[/tex]
C. [tex]f(1) = 2[/tex]
D. [tex]f(1) = 3[/tex]


Sagot :

To find the value of [tex]\( f(1) \)[/tex] for the given piecewise function, we need to carefully evaluate the function at [tex]\( x = 1 \)[/tex].

Given the piecewise function:
[tex]\[ f(x) = \begin{cases} x^2 + 1, & -4 \leq x < 1 \\ -x^2, & 1 \leq x < 2 \\ 3x, & x \geq 2 \end{cases} \][/tex]

1. First, identify which interval [tex]\( x = 1 \)[/tex] falls into:
- The interval [tex]\([-4 \leq x < 1)\)[/tex] does not include [tex]\(1\)[/tex], since it is strictly less than 1.
- The interval [tex]\([1 \leq x < 2)\)[/tex] does include [tex]\(1\)[/tex], because [tex]\(1\)[/tex] lies within the range [tex]\(1 \leq x < 2\)[/tex].
- The interval [tex]\([x \geq 2)\)[/tex] does not include [tex]\(1\)[/tex], since [tex]\(1\)[/tex] is less than [tex]\(2\)[/tex].

2. [tex]\( x = 1 \)[/tex] falls into the interval [tex]\( [1 \leq x < 2) \)[/tex]. For this interval, the function is defined as:
[tex]\[ f(x) = -x^2 \][/tex]

3. Now substitute [tex]\( x = 1 \)[/tex] into this function definition:
[tex]\[ f(1) = -(1)^2 = -1 \][/tex]

Therefore, the value of [tex]\( f(1) \)[/tex] is [tex]\(\boxed{-1}\)[/tex].
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