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Say we have this function: $g(x) = 1+ \frac{1}{x+1}$ for $x>-1$.

The question asks to solve for $g(x) = g^{-1}(x)$. I intuitively thought to first calculate the inverse function's equation which turned out to be $y=\frac{-x+2}{x-1}$ and then to simply equate it with the original function (which I thought was the normal way to find POI's of functions). However, while this did produce an answer ($x=0,-2$) they were not correct (the answer is $x=\sqrt{2}$.Could I have some direction on this please.

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Your method is correct, but you made a mistake when finding the inverse of $g$. To see why, note that $g(0)=2$ but $$ -\frac{(2)+2}{(2)-1} = -4 \ne 0. $$ It should be $$ g^{-1}(x) = \frac{2-x}{x-1}. $$ Let's verify that this is correct. Note that $$ (g\circ g^{-1})(x) = 1+\frac{1}{\frac{2-x}{x-1}+1} = 1+\frac{1}{\frac{1}{x-1}} = 1+(x-1)=x $$ and $$ (g^{-1}\circ g(x) = \frac{2-(1+\frac{1}{x+1})}{(1+\frac{1}{x+1})-1} = \frac{1-\frac{1}{x+1}}{\frac{1}{x+1}} = \frac{1}{\frac{1}{x+1}} -1=(x+1)-1=x. $$


Continuing with the solution, we need to solve $g(x)=g^{-1}(x)$. That is: \begin{align*} 1+\frac{1}{x+1} &= \frac{2-x}{x-1} \\ (x+1)(x-1) + (x-1) &= (x+1)(2-x) \\ (x-1)(x+2) &= (x+1)(2-x) \\ x^2+x-2 &= -x^2+x+2 \\ 2x^2-4&=0 \\ x^2 &= 2 \\ x &= \pm\sqrt{2} \end{align*}

Since the domain excludes $x=-\sqrt{2}$, we are left with only $\sqrt{2}$.

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  • $\begingroup$ argh! i made a typo when i was putting this online! on my paper i have it as what i have corrected it to. i am very sorry for wasting your time, but your answer still is insightful $\endgroup$ – John Hon Aug 19 '17 at 14:50
  • $\begingroup$ @JohnHon No problem. I've added more information. $\endgroup$ – John Griffin Aug 19 '17 at 14:59
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HINT: you have to solve $$1+\frac{1}{x+1}=\frac{1}{x-1}-1$$ and this is equivalent to $$2(x^2-2)=0$$

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