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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]
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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