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If f(x) = 2/3x - 6, find f^-1. Justify your answer using the composition of functions f(f^-1(x))=f^-1(f(x))

Sagot :

MrRoyal

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]

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