What is the range of the function $f(x)=\frac{4x(x^2+1)}{x^2+(x^2+1)^2}$?

Question

What is the range of the function $$ f(x)=\frac{4x(x^2+1)}{x^2+(x^2+1)^2} $$ with $x\ge0$?

I can't find a right method to solve this question. Any help would be appreciated.

Answer 1

It's obvious that $f(x)\geq0.$

Let $x>0$ and $x+\frac{1}{x}=2t$.

Thus, by AM-GM $$2t=x+\frac{1}{x}\geq2\sqrt{x\cdot\frac{1}{x}}=2,$$ which says $t\geq1$.

The equality occurs for $x=1$.

Thus, by AM-GM again $$f(x)=\frac{4}{\frac{x}{x^2+1}+\frac{x^2+1}{x}}=\frac{4}{\frac{1}{2t}+2t}=\frac{4}{\frac{1}{2t}+\frac{t}{2}+1.5t}\leq\frac{4}{2\sqrt{\frac{1}{2t}\cdot\frac{t}{2}}+1.5}=\frac{8}{5}.$$ Since $f$ is a continuous function, we got the answer: $$\left[0,\frac{8}{5}\right]$$

Answer 2

The derivative of $f(x)$ is :

$$f'(x) = - \frac{4(x^6-1)}{(x^4 + 3x^2 +1)^2}$$

I'll leave the calculation of it to you, since it's a simple fraction derivative application, just make sure to be careful to not make any numerical mistakes, as its calculation is very easy, by applying the fraction differentiation rule :

$$\bigg( \frac{f(x)}{g(x)}\bigg)' = \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}$$

Obviously, the denominator of the fraction is always positive, since : $(x^4 + 3x^2 + 1)^2 > 0$ .

This means that the sign of the derivative depends on the sign of the numerator, specifically :

$$4(x^6-1) > 0 \Rightarrow x^6 - 1 > 0 \Rightarrow x>1 \space \text{or} \space x < -1$$

But on the exercise it's mentioned that we're exploring over $x \geq 0$, so this means that we will only take into account the solution of the inequality : $x>1$.

This means that at $x=1$ your function changes sign, which by elementary calculus rules this means that this is a global maximum for your function, which will specifically be :

$$f(1) = \frac{8}{5}$$

So, in the domain $x \in \mathbb [0,+ \infty)$, we only need to check its behavior at infinite, to determine whether we'll need another value :

$$\lim_{x \to \infty} \frac{4x(x^2+1)}{x^2+(x^2+1)^2}=0$$

which means that the function asymptotically tends to $0$ as $x$ becomes very large.

Also note that $f(0) = 0$.

These are enough to conclude that the range of the values of the function for $x\geq 0$, are :

$$f([0, + \infty))= [0, 8/5]$$

Answer 3

For $u\in]-\frac\pi2,\frac\pi2[$ let's substitute $x=\tan(u)$

$f(x)=\dfrac{4x(1+x^2)}{x^2+(1+x^2)^2}=\dfrac{\frac{4\sin(u)}{\cos(u)^3}}{\frac{\sin(u)^2\cos(u)^2+1}{\cos(u)^4}}=\dfrac{4\sin(u)\cos(u)}{1+\sin(u)^2\cos(u)^2}=\dfrac{2\sin(2u)}{1+\frac 14\sin(2u)^2}=\dfrac{8s}{4+s^2}$

With $s=\sin(2u)\in]-1,1[$.

Let's set $g(x)=\dfrac{8x}{4+x^2}$

$g'(x)=\dfrac{-8(x-2)(x+2)}{(4+x^2)^2}>0$ on the given interval so $g\nearrow$.

The extrema are thus given for the bounds of the interval : $g(1)=\dfrac 85$ and $g(-1)=-\dfrac 85$.

Since $g$ is continuous all values in $]-\frac 85,\frac 85[$ are reached.

The substitution between $x$ and $u$ being bijective, we conclude the same for the range of $f$.