Show that $f(x) = { 1\over x^2 + \ln^2x }$ is bounded

Question

How do i show that $f(x) = { 1\over x^2 + \ln^2x}$ is bounded?

Let $g(x) = x^2 + \ln^2x$.

The domain of $f(x)$ is $x > 0$, for the denominator we have that $x^2 > 0$ and $\ln^2x > 0$ hence ${ 1 \over g(x) } > 0$ for any $x$. Choosing large values for $x$ we have that $g(x)$ tends to infinity and hence $1 \over g(x)$ tends to $0$. Based on that observations we have that $f(x) > 0$.

But how do I show that the function has an upper bound? After plotting the graph it's clear that it exists, but I couldn't find a way to show that analytically. I'm not supposed to use derivatives but even if i could it gives that:

$$ f'(x) = -\frac{2(x^2 + \ln(x))}{x(x^2 + \ln^2(x))^2} $$

Then i need to solve for $f'(x) = 0$, which I assume is solvable only by using Lambert's W-function.

Answer 1

You can define a new continuous function $$ f(x)=\begin{cases} \dfrac{1}{x^2+(\ln x)^2} & x>0 \\[4px] 0 & x=0 \end{cases} $$ and consider that $f(1)=1$. Since $$ \lim_{x\to\infty}f(x)=0 $$ there exists $a>0$ so that $f(x)<1$ for every $x>a$.

The function $f$, restricted to the interval $[0,a]$, is continuous, so it assumes a maximum value $M\ge1$, which is also the maximum for $f$ over $(0,\infty)$.

Thus $0\le f(x)\le M$ for every $x\in(0,\infty)$ and so the function is bounded.

Answer 2

We have that

• $\lim_{x\to 0} f(x)=0$

• $\lim_{x\to \infty} f(x)=0$

• $x^2+\log^2 x>0$

therefore for EVT $f(x)$ has maximum on the domain $x>0$ and therefore $$0<f(x)\le M$$

Answer 3

We have $$ g'(x)=2x+2\frac{\log(x)}x $$ and $$ g''(x)=2\frac{1+x^2-\log(x)}{x^2} $$ Note that $g'\left(\frac12\right)=1-4\log(2)\lt0$ and $g'(1)=2\gt0$. Therefore, for some $x_0\in\left[\frac12,1\right]$, we must have $g'(x_0)=0$. Furthermore, $g''(x)\gt0$ for all $x$. Thus, for all $x$, $g(x)\ge g(x_0)$ and therefore, $$ f(x)\le\frac1{g(x_0)} $$

Answer 4

If $x\geq 1/2$, then $x^2+\ln(x)^2>x^2 >1/4.$

If $x \leq 1/2,$ then $x^2+\ln(x)^2> (-\ln x)^2 >\ln(2)^2>1/4.$

So your $g(x) \geq 1/4$ and $f(x) \leq 4.$