SOLUTION: Let 𝑓 be the function with correspondence rule 𝑓(𝑥) =−3−2𝑥/𝑥+3, determine the inverse function 𝑓^−1 and its domain.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Let 𝑓 be the function with correspondence rule 𝑓(𝑥) =−3−2𝑥/𝑥+3, determine the inverse function 𝑓^−1 and its domain.      Log On


   

Question 1194863: Let 𝑓 be the function with correspondence rule 𝑓(𝑥) =−3−2𝑥/𝑥+3, determine the inverse function 𝑓^−1 and its domain.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


One standard way to find the inverse of a function is to switch the x and y and solve for the new y.

I would start be rewriting the given function in a more standard form: y=%28-2x-3%29%2F%28x%2B3%29

Now switch and solve for the new y.

x=%28-2y-3%29%2F%28y%2B3%29
x%28y%2B3%29=-2y-3
xy%2B3x=-2y-3
xy%2B2y=-3x-3
y%28x%2B2%29=-3x-3
y=%28-3x-3%29%2F%28x%2B2%29

ANSWER: The inverse function is y=%28-3x-3%29%2F%28x%2B2%29

The domain of the function is all real numbers except any that make the denominator zero: (-infinity,-2) U (-2,infinity)

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A couple of added notes about this problem....

(1) The horizontal asymptote of the given function is the value as x approaches positive or negative infinity, which is (-2x)/(x) = -2; the single value not in the domain of the inverse is the y value of the horizontal asymptote of the given function. (The horizontal asymptote of the given function is the vertical asymptote of the inverse function.)

(2) On competitive math contests, this problem of finding the inverse of a rational function of the form (ax+b)/(cx+d) is quite common. The computations involved in finding the inverse are always similar to those above for this particular example. And the result of those computations always shows the same pattern:

The inverse of the rational function (ax+b)/(cx+d) is (-dx+b)/(cx-a).

The pattern is that the "b" and "cx" stay where they are, while the "a" and "d" switch places and both change sign.

Using that pattern for the given example...

(-2x-3)/(x+3) --> (-3x-3)/(x+2)