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$\begingroup$

$g(x) =\frac{3-5x}{5-x}$

I know that I am retarded for not being able to do this on my own. Algebraically you just have to solve for x. Even so, I can't do it.

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    You're not retarded if can't solve something. Generally in mathematics you need a lot of practice when learning something so that there will come a time when you can see that problem as "easy", but the thing is that one usually tends to forget how many hours were spent practicing and solving exercises to get to that stage. Obviously there are things that cannot be considered easy no matter how much you practice or study.2010-11-20
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    Suggestion: Begin by multiply both sides by $5-x$ (which you can do provided that $5-x\neq 0$). Then transform your equation into the form $(cg(x)+d)x=(ag(x)+b)$ by adding or subtracting suitable terms to both sides. Finally solve for $x=\dfrac{ag(x)+b}{cg(x)+d}$. This expression is valid for $cg(x)+d\neq 0$. The inverse function is $g(x)\mapsto \dfrac{ag(x)+b}{cg(x)+d}$2010-11-20
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    I edited your title; no need to beat down yourself in front of us.2010-11-20

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