Verify/Check the Inverse...

Question

For g(x)=1/3x-2, verify/check the inverse for g(x). 

 

Find g^-1(x). Graph both g(x) & g^-1(x) to show that they have symmetry with respect to y=x.

Answer 1

To find the inverse of a function

1) replace g(x) with y

2) swap x and y in the equation

3) solve for y

4) replace y with g^-1(x)

This final answer is the inverse function

 

If your original function is g(x) = 1/(3x-2)

 

For graphing purposes note we cannot have 3x-2=0 so x≠2/3

There is a vertical asymptote at x = 2/3

For horizontal asymptote look at degree of numerator and denominator

degree numerator = 0 since 1 = 1x⁰,  degree denominator = 1

Since degree of the denominator is greater than degree of numerator there is

a horizontal asymptote at y=0

 

Let's find the inverse function g^-1(x)

1)  y = 1/(3x-2)

2) x = 1/(3y-2)

3) x(3y-2) = 1 

    3y-2=1/x

    3y = 1/x + 2

     y = 1/(3x) + 2/3

4)  g^-1(x) = 1/3x + 2/3

 

For graphing purposes lets put this into a single fraction

g^-1(x) = 1/3x + 2/3  = 1/3x + 2x/3x= (2x + 1)/3x

 

Again, we cannot divide by zero so there is a vertical asymptote at x=0

There is a horizontal asymptote at y = 2/3 since the degree numerator = degree denominator

and dividing the coefficients of the x terms gives 2/3

 

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If original function is g(x) = 1/3x - 2

If we put this into a single fraction we have g(x) = (1-6x)/(3x) or g(x) = (-6x+1)/(3x)

 

For graphing note there is a vertical asymptote at x=0 since we cannot divide by 0

The degree of the denominator is equal to degree of the numerator so divide the coefficients

giving -6/3 = -2.. there is a horizontal asymptote at y = -2

 

Let's find the inverse function g^-1(x)

1) y = 1/3x - 2

2) x = 1/3y - 2

3) 1/3y = x + 2

    1 = 3y(x+2)

    y = 1/(3(x+2))

    y = 1/(3x+6)

4) g^-1(x) = 1/(3x+6)

 

Note when graphing this that x≠-2   since  3(-2) + 6 = 0 and we can't divide by 0

There is a vertical asymptote at x=-2

For horizontal asymptote look at the degree of exponents on numerator and denominator

Since the degree of the denominator is greater than the numerator there is a horizontal

asymptote at y=0