Certainly! Let's tackle each part of this problem one step at a time.
### Part a) Find [tex]\( f(3) \)[/tex]
Given the function:
[tex]\[ f(x) = \frac{x^2 - 3}{2} \][/tex]
To find [tex]\( f(3) \)[/tex], substitute [tex]\( x = 3 \)[/tex] into the function:
[tex]\[ f(3) = \frac{3^2 - 3}{2} \][/tex]
[tex]\[ f(3) = \frac{9 - 3}{2} \][/tex]
[tex]\[ f(3) = \frac{6}{2} \][/tex]
[tex]\[ f(3) = 3 \][/tex]
So, [tex]\( f(3) = 3 \)[/tex].
### Part b) Find [tex]\( f^{-1}(x) \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we start with the equation for [tex]\( f(x) \)[/tex] and solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
Start with:
[tex]\[ y = f(x) = \frac{x^2 - 3}{2} \][/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{x^2 - 3}{2} \][/tex]
[tex]\[ 2y = x^2 - 3 \][/tex]
[tex]\[ x^2 = 2y + 3 \][/tex]
[tex]\[ x = \pm \sqrt{2y + 3} \][/tex]
So, the inverse function [tex]\( f^{-1}(x) \)[/tex] can be written as:
[tex]\[ f^{-1}(x) = \pm \sqrt{2x + 3} \][/tex]
### Part c) Find [tex]\( f^{-1}(11) \)[/tex]
Using the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \pm \sqrt{2x + 3} \][/tex]
Substitute [tex]\( x = 11 \)[/tex] into the inverse function:
[tex]\[ f^{-1}(11) = \pm \sqrt{2 \cdot 11 + 3} \][/tex]
[tex]\[ f^{-1}(11) = \pm \sqrt{22 + 3} \][/tex]
[tex]\[ f^{-1}(11) = \pm \sqrt{25} \][/tex]
[tex]\[ f^{-1}(11) = \pm 5 \][/tex]
Therefore, the solutions are:
[tex]\[ f^{-1}(11) = -5, 5 \][/tex]
### Summary
- a) [tex]\( f(3) = 3 \)[/tex]
- b) [tex]\( f^{-1}(x) = \pm \sqrt{2x + 3} \)[/tex]
- c) [tex]\( f^{-1}(11) = -5, 5 \)[/tex]
