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This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.

It is easy to find $f^{-1}$ and solve this like below ... $$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})^2\\4y^2+x^2-4xy=x^2+4\\4y^2-4=4xy\\x=\frac{y^2-1}{y}=y-\frac{1}{y} \\ \to f^{-1}(x)=x-\frac{1}{x}\\ f^{-1}(x)+f^{-1}(\frac{1}{x})=x-\frac{1}{x}+(\frac{1}{x}-\frac{1}{\frac{1}{x}})=0\\$$ And now my question is ... is there an other method to solve this question ?

I was thinking about $f(x)f(-x)=1$ but I can't go anymore ... Any hint ,or other Idea ? Thanks in advanced.

  • 0
    $f^{-1}(x)$ is not defined for $x<0$. – 2017-01-10
  • 1
    Yes . Because $f(x) >0$. – 2017-01-10

4 Answers 4

10

Perhaps your observation can be made to work (note $f(x) \neq 0$): \begin{align*} f(x)f(-x) & =1\\ f(-x) & = \frac{1}{f(x)}\\ f^{-1}(f(-x)) & = f^{-1}\left(\frac{1}{f(x)}\right)\\ -x & = f^{-1}\left(\frac{1}{f(x)}\right)\\ f^{-1}\left(\frac{1}{f(x)}\right)+x & = 0\\ f^{-1}\left(\frac{1}{f(x)}\right)+f^{-1}(f(x)) & = 0\\ f^{-1}\left(\frac{1}{y}\right)+f^{-1}(y) & = 0 \end{align*}

3

The short version:

$$f(x)f(-x)=1\implies f^{-1}(f(x))+f^{-1}\left(\frac1{f(x)}\right)=f^{-1}(f(x))+f^{-1}\left(f(-x)\right)=0.$$

2

Disclaimer: Just offering a geometrical perspective. OP's solution and Anurag's solutions are much more elegant.

If we are given the quadratic equation $$g(z)=az^2+bz+c,$$

we know that the bigger root is $$\frac{-b+\sqrt{b^2-4ac}}{2a}$$

View $x$ as a parameter, when we are given $x$, we let $b = -x$, $a = 1$, $c=-1$.

The function $f(x)$ returns the bigger root of $$g(z)=z^2-xz-1$$

This quadratic equation has vertical intercept $-1$ and it is symmetric about $z=\frac{x}{2}$.

Suppose the bigger root is $y_b$ and the smaller root is $y_s$.

We know that $y_by_s=-1$, the smaller root $y_s=\frac{-1}{y_b}$

Hence this quadratic equation passes through $(0,-1), (y_b,0)$ and $(-\frac{1}{y_b},0)$

Now let's try to understand what does $f^{-1}(y)$ do geometricallly. $f^{-1}(y)$ construct the quadratic equation that passes through $(0,-1), (y,0)$ and $(-\frac1y, 0)$, (let me call this quadratic curve $Q_1$)look for the symmetrical line for the quadratic curve $z=\frac{x}{2}$ and returns $x$.

$$x=f^{-1}(y)$$

(actually we know that $x=y-\frac{1}{y}$)

Now let's try to understand what does $f^{-1}(\frac1y)$ do geometrically.$f^{-1}(\frac1y)$ construct the quadratic equation that passes through $(0,-1),(\frac1y,0)$ and $(-y,0)$, (let me call this quadratic curve $Q_2$), look for the symmetrical line for the quadratic curve $z = \frac{\hat{x}}{2}$ and returns $\hat{x}$

$$\hat{x}=f^{-1}(\frac1y)$$

(again, actually we know that $\hat{x}=\frac1y-y$)

$Q_2$ is actually just the reflection of $Q_1$ with reflection axis $z=0$.

Hence $$\hat{x}=-x$$

$$x+\hat{x}=0$$

$$f^{-1}(y)+f^{-1}(\frac1y)=0$$

2

$f(2\sinh\theta) = \frac{1}{2}\left(2\sinh\theta+2\cosh\theta\right) = e^\theta $ implies that the range of $f$ is $\mathbb{R}^+$ and $$ f(x) = \exp\text{arcsinh }\frac{x}{2} \to \\ \qquad f^{-1}(x)=2\sinh\ln x=x-\frac{1}{x}\tag{1}$$ so $ f^{-1}(x)+f^{-1}\left(\frac{1}{x}\right) = \color{red}{0}$ for any $x\in\mathbb{R}^+$.