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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 determine the value of [tex]\( f(1) \)[/tex] for the piecewise defined function [tex]\( f(x) \)[/tex], we need to examine which part of the piecewise function applies when [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]

We look at the domain of each piece:

1. [tex]\( x^2 + 1 \)[/tex] is defined for [tex]\( -4 \leq x < 1 \)[/tex].
2. [tex]\( -x^2 \)[/tex] is defined for [tex]\( 1 \leq x < 2 \)[/tex].
3. [tex]\( 3x \)[/tex] is defined for [tex]\( x \geq 2 \)[/tex].

Since [tex]\( x = 1 \)[/tex] falls into the interval [tex]\( 1 \leq x < 2 \)[/tex], we use the second part of the piecewise function, which is [tex]\( -x^2 \)[/tex].

Now, we calculate [tex]\( f(1) \)[/tex]:

[tex]\[ f(1) = -1^2 = -1 \][/tex]

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

So the correct answer is:
[tex]\[ f(1) = -1 \][/tex]
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