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Question

find the inverse of 2x-3/x+4

what is the domain of f and it's range of f^-1

Answer 1

The given fuction f(x) = 2x -3/x+4

The inverse fuction of y = f(x)

y = 2x - 3/x+4

2x - 3 = y (x+4)

2x - 3 = xy + 4y

2x - xy = 4y +3

x ( 2 - y) = 4y +3

x = ( 4y+3)/(2 - y)

Therefore f^-1(x) = ( 4x+3)/(2 - x)

The domain f is 2x - 3/x+4

(x+4)  is not equal to 0

Therefore x is not equal to ( - 4 )

Domain  = X - belongs to { -∞,∞ } - { -4}

The range of the f^-1(x) is ( 4x+3)/(2-x) not equal to zero

4x+3 =0

x = -3/4

Therefore Range = { -3/4 }.

 

Comments

The inverse function is f ^{- 1}(x) = (4x + 3) / (2 - x).

Let y = f ^{- 1}(x) = (4x + 3) / (2 - x).

In the polynomial function, x represents the domain and y represent the range.

To find the range of the function solve for x.

y = (4x + 3) / (2 - x)

y(2 - x) = (4x + 3)

2y - xy = 4x + 3

- xy - 4x = 3 - 2y

x(- y - 4) = 3 - 2y

x = (3 - 2y) / (- y - 4)

x = (2y - 3) / (y + 4)

Therefore, y + 4 ≠ 0 ⇒ y ≠ - 4.

The range of the f ^{- 1}(x) = (4x + 3) / (2 - x) is {(- ∞,∞) - (- 4) }.

Secons method :

For the rational function, a vertical asymptote, is at some x value, as you should expect from the domain restriction and a horizontal asymptote, is at some y value, as you should expect from the range restriction.

To find the vertical asymptote of rational function, denominator D(x) = 0.

Find the horizontal asymptote :

The degree of N(x) = degree of D(x), so y = a_n/b_m (ratio of the leading coefficients) as a horizontal asymptote. y = 4/(-1) = - 4.

Range : ( (- ∞,∞) - (- 4) ).