Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $$f(x) = \frac{1}{2x+1}$$ and the inverse f is $$f^{-1}(x) = \frac{1- x}{2x}$$ How do I find the two possible values of $x$ when $f(x) = f^{-1}(x)$?

share|improve this question
1  
Please use $\LaTeX$ formatting when posting questions. Also tag homework as such. –  Your Ad Here Oct 16 '12 at 20:00
add comment

2 Answers

When $f(x) = f^{-1} (x)$, we have

$$ \frac{1}{2x+1} = \frac{1-x}{2x} \\ \implies 2x = (1-x)(2x+1) \\ \implies 2x = -2x^2 + x +1 $$

This is a quadratic and will have two roots. Can you use the quadratic formula?

share|improve this answer
    
Yeah, i take it i'd move the 2x to the other side to make -x, so a = -2, b = -1 and c = 1? –  pieltd Oct 16 '12 at 20:06
    
Yes, correct.... –  Ganesh Oct 16 '12 at 20:15
add comment

Suppose that for some $x$, $f(x) = f^{-1}(x)$, then

\begin{align} f(x) = \frac{1}{2x + 1} & = \frac{1-x}{2x} = f^{-1}(x). \end{align}

Cross multiplying, we have $$2x = (1-x)(2x+1)$$

which, upon expansion and rearrangement, we have

$$2x^2 + x - 1 = 0.$$

To solve that equation, we may either use the quadratic formula, or notice that we may factor the quadratic to

$$(2x - 1)(x + 1) = 0$$

and so we see that $x = 1/2$ and $x = -1$ are both solutions to the quadratic. It is important to check that these are indeed solutions to our equation, since we do have two rational functions and they may very well be undefined on these points. Luckily, they do not cause our denominators to be $0$ in either case, both of these roots are solutions to our original problem.

Note. To solve the problem, we assumed that there was indeed some $x$ such that they are equal. This is not necessarily rigorous, per se, but it is fine for our purpose here.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.