Show that $f:[0,\infty)\to [0,\infty)$ with $f(x)=x^2\log(1+x^2)$ is bijective.

Question

Define $f:[0,\infty)\to [0,\infty)$ with $f(x)=x^2\log(1+x^2).$ How can I show that $f$ is bijective and how can I get the derivative of the inverse function in $\log(2)$?

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Answer 1

We compute the derivative \begin{align*} f'(x) =& 2x \log (1+x^2) + x^2 \frac{2x}{1+x^2} \\ =& 2x \left( \log(1+x^2) + \frac{x^2}{1+x^2} \right) \geq 0 \end{align*} and we only get equality when $x=0$. Hence, the function is increasing, and therefore it is a bijection.

Now, to calculate the derivative of the inverse, consider the following \begin{align*} y=&f^{-1}(x) \\ f(y) =& x \\ f'(y)y' =& 1 \\ y' =& \frac{1}{f'(y)} \\ y' =& \frac{1}{f'(f^{-1}(x))} \end{align*} So, evaluating the inverse derivative in $\log(2)$ means first finding that $x$ for which $f(x)= \log(2)$. We can easily see that $x=1$ suffices. Now $f'(1) = 2 \left( \log(2) + \frac{1}{2} \right)$, and hence $(f^{-1})'(\log(2)) = \frac{1}{2 \left( \log(2) + \frac{1}{2} \right)}$.

Answer 2

$$f'(x)=2x\log(1+x^2)+\frac{2x^3}{1+x^2}>0\;\;,\;\;\forall\,x>0$$

and from the above it follows the function is monotonic increasing and thus $\;1-1\;$ , and since it is differentiable and thus continuous, and $\;f(0)=0\;,\;\;\lim\limits_{x\to\infty} f(x)=\infty\;$ , it also is onto $\;[0,\infty)\;$ .

From the above, it follows it is invertible, and from the theorem on the derivative of the inverse function we get that

$$\left(f^{-1}\right)'(\log2)=\frac1{f'(a)}\;,\;\;\text{when }\;a\;\;\text{is such that}\;\;f(a)=\log2$$

Answer 3

By calculating the derivative of the function, $$f'(x)=2xln(x^2+1) + \frac{2x^3}{x^2+1}$$ you find that $f$ is a monotonic increasing function, because $f'(x)$ is always positive for $0< x<\infty$. But if $f$ is monotonic, it is also injective. Then consider $y s.t. 0<y<\infty$. You can see that $\forall y$, $f$ is definitely greater than $y$, and that is because $\lim_{x\to 0} f(x) = \infty$. And so, using the Intermediate values theorem you get that $\forall y\ \exists x\ s.t. f(x)=y$. This means that $f$ is bijective.

To calculate the derivative of the inverse in $log(2)$ use the formula $$g'(y)=\frac{1}{f'(x)}$$ where $g$ is the inverse of $f$. You can see easily that $f(1)=log(2)$ so the value of the inverse's derivative when $y=log(2)$ is $\frac{1}{2log(2)+1}$.