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Answer:
[tex]f^{-1}(x)= \frac{3}{2}(x+6)[/tex]
[tex]f(f^{-1}(x)) = f^{-1}(f(x))= x[/tex]
Step-by-step explanation:
Given
[tex]f(x) = \frac{2}{3}x - 6[/tex]
Required
Determine [tex]f^{-1}(x)[/tex]
[tex]f(x) = \frac{2}{3}x - 6[/tex]
Represent f(x) as y
[tex]y=\frac{2}{3}x - 6[/tex]
Swap the positions of x and y
[tex]x=\frac{2}{3}y - 6[/tex]
Add 6 to both sides
[tex]x+6=\frac{2}{3}y - 6+6[/tex]
[tex]x+6=\frac{2}{3}y[/tex]
Multiply both sides by 3
[tex]3(x+6)=\frac{2}{3}y * 3[/tex]
[tex]3(x+6)=2y[/tex]
Divide both sides by 2
[tex]\frac{3}{2}(x+6)=\frac{2y}{2}[/tex]
[tex]\frac{3}{2}(x+6)=y[/tex]
[tex]y= \frac{3}{2}(x+6)[/tex]
Replace y with [tex]f^{-1}(x)[/tex]
[tex]f^{-1}(x)= \frac{3}{2}(x+6)[/tex]
Justify the result:
First, we solve for [tex]f(f^{-1}(x))[/tex]
[tex]f(x) = \frac{2}{3}x - 6[/tex]
[tex]f(x) = \frac{2}{3}x - 6[/tex] becomes
[tex]f(f^{-1}(x)) = \frac{2}{3}x - 6[/tex]
Substitute [tex]\frac{3}{2}(x + 6)[/tex] for x
[tex]f(f^{-1}(x)) = \frac{2}{3}(\frac{3}{2}(x + 6)) - 6[/tex]
[tex]f(f^{-1}(x)) = \frac{2*3}{3*2}(x + 6)- 6[/tex]
[tex]f(f^{-1}(x)) = \frac{6}{6}(x + 6)- 6[/tex]
[tex]f(f^{-1}(x)) = 1*(x + 6)- 6[/tex]
[tex]f(f^{-1}(x)) = x + 6- 6[/tex]
[tex]f(f^{-1}(x)) = x[/tex]
Next, we solve [tex]f^{-1}(f(x))[/tex]
[tex]f^{-1}(x)= \frac{3}{2}(x+6)[/tex]
[tex]f^{-1}(x)= \frac{3}{2}(x+6)[/tex] becomes
[tex]f^{-1}(f(x))= \frac{3}{2}(x+6)[/tex]
Substitute [tex]\frac{2}{3}x - 6[/tex] for x
[tex]f^{-1}(f(x))= \frac{3}{2}(\frac{2}{3}x) - 6+6[/tex]
[tex]f^{-1}(f(x))= \frac{3}{2}(\frac{2}{3}x)[/tex]
[tex]f^{-1}(f(x))= \frac{2*3}{3*2}x[/tex]
[tex]f^{-1}(f(x))= \frac{6}{6}x[/tex]
[tex]f^{-1}(f(x))= 1*x[/tex]
[tex]f^{-1}(f(x))= x[/tex]
Hence:
[tex]f(f^{-1}(x)) = f^{-1}(f(x))= x[/tex]