What is the value of [tex]\( f(1) \)[/tex] for the piecewise defined function below?

[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]



Answer :

To determine the value of [tex]\( f(1) \)[/tex] for the given piecewise function, we need to see which piece of the function applies when [tex]\( x \)[/tex] is equal to 1.

The piecewise function is defined as follows:
[tex]\[ f(x) = \begin{cases} x^2 + 1, & \text{for } -4 \leq x < 1 \\ -x^2, & \text{for } 1 \leq x < 2 \\ 3x, & \text{for } x \geq 2 \end{cases} \][/tex]

We are interested in the value of [tex]\( f(1) \)[/tex]. Let's locate which piece of the function covers [tex]\( x = 1 \)[/tex]:

1. The first piece [tex]\( x^2 + 1 \)[/tex] applies for [tex]\( -4 \leq x < 1 \)[/tex]. Since 1 is not within this interval (it is the endpoint but not less than 1), this piece does not apply.
2. The second piece [tex]\( -x^2 \)[/tex] applies for [tex]\( 1 \leq x < 2 \)[/tex]. Since [tex]\( x = 1 \)[/tex] falls within this interval, we use this piece.
3. The third piece [tex]\( 3x \)[/tex] applies for [tex]\( x \geq 2 \)[/tex]. Since 1 is not greater than or equal to 2, this piece does not apply.

Now, evaluate the function using the relevant piece:
[tex]\[ f(x) = -x^2 \][/tex]
At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -(1)^2 = -1 \][/tex]

The value of [tex]\( f(1) \)[/tex] is [tex]\(-1\)[/tex], which matches the provided answer.

Therefore, the correct option is:
[tex]\[ f(1) = -1 \][/tex]