Certainly! Let's find the inverse function of [tex]\( f(x) = 2x - 1 \)[/tex] and evaluate it at 3.
1. Define the function:
[tex]\( f(x) = 2x - 1 \)[/tex]
2. Set [tex]\( y \)[/tex] to be equal to [tex]\( f(x) \)[/tex]:
[tex]\( y = 2x - 1 \)[/tex]
3. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y = 2x - 1
\][/tex]
[tex]\[
y + 1 = 2x
\][/tex]
[tex]\[
x = \frac{y + 1}{2}
\][/tex]
Therefore, the inverse function [tex]\( f^{-1}(y) \)[/tex] is:
[tex]\[
f^{-1}(y) = \frac{y + 1}{2}
\][/tex]
4. Evaluate the inverse function at [tex]\( y = 3 \)[/tex]:
[tex]\[
f^{-1}(3) = \frac{3 + 1}{2} = \frac{4}{2} = 2
\][/tex]
Hence, [tex]\( f^{-1}(3) = 2 \)[/tex].
The correct answer is:
(B) 2
