To find the boundedness of a given function

[brotherbobby]

Homework Statement

Find the boundedness of the function $\quad\boldsymbol{f(x)=\dfrac{x^2}{x^4+1}}$

Relevant Equations

A function $f(x)$ is said to be bounded above if one can find a number $M$ for which $\vert f(x)\vert\le M$.

Likewise, a function $f(x)$ is said to be bounded below if one can find a number $N$ for which $\vert f(x)\vert\ge N$.

Given an equation quadratic in $x$, $ax^2+bx+c=0$, for all real values of $x$, the discriminant $\mathscr{D}=b^2-4ac\ge 0$

$\small{\texttt{(I could solve the for the upper limit explicitly.}}$
$\small{\texttt{However, not the same for the lower limit, except via inspection.)}}$

I copy and paste the the problem as it appeared in the text.

$\rm(I)$ : $\texttt{The domain :}$

The domain of the function is clearly all real values of $x$. Hence $\rm{D}(f(x)) = x\in \mathbb{R}$

$\rm(II)$ : $\texttt{The range and boundedness :}$

If $f(x)=\frac{x^2}{x^4+1}=y\Rightarrow \color{blue}{y\left( x^2 \right)^2-x^2+y=0}\Rightarrow\text{Discriminant}\;\mathscr{D}=1-4y^2\ge 0\Rightarrow -\frac{1}{2}\le y \le \frac{1}{2}$

However, $y=\frac{x^2}{x^4+1}\ge 0$. Hence the bounds of $\boxed{f(x)\in\left[0, \frac{1}{2}\right]}\quad\Large{\color{green}\checkmark}$.
The answer matches the one in the text, shown alongside.


Doubt : $\small{\texttt{I could find the lower limit of y above through inspection.}}$
$\small{\texttt{However, what if I tried to find it explicitly? I run into problems.}}$

Let's return to the quadratic in $\color{blue}{\text{blue}}$ above : $\color{blue}{y\left( x^2 \right)^2-x^2+y=0}$.

We have $x^2\ge 0$ besides being real. Hence, solving for $x^2=\frac{1\pm\sqrt{1-4y^2}}{2}\ge 0\Rightarrow 1\pm\sqrt{1-4y^2}\ge 0\Rightarrow \pm\sqrt{1-4y^2} \ge -1\Rightarrow 1-4y^2\ge 1 \Rightarrow 4y^2\le 0\Rightarrow y=0\color{red}{\huge\times}$

Request : A hint as to where am going wrong above. I should be able to show, explicitly, that $f(x)\in\left[ 0, \frac{1}{2}\right]$

 

[PeroK]

When you write $1 \pm \sqrt{1 - 4y^2} \ge 0$, that is two separate conditions: $1 + \sqrt{1 - 4y^2} \ge 0$ and $1 - \sqrt{1 - 4y^2} \ge 0$. Neither implies $y = 0$.

Also, you know $y \ge 0$ from the definition of the function.

 

[WWGD]

You know that outside of $[-1,1]$, that $x^4>x^2$. The issue then, is to see what happens in $[-1,1]$. Can you think of what to use to figure it out, Brother Bobby?

 

[brotherbobby]

Also, you know y≥0 from the definition of the function.

Yes, I said as much in my solution attempt in post #1. However, I am seeking to do this problem without the inspection of function $y$ itself, but finding out that $y\ge 0$ explicitly.

When you write 1±1−4y2≥0, that is two separate conditions: 1+1−4y2≥0 and 1−1−4y2≥0. Neither implies y=0.

(1) Let me take the first condition :

$1+\sqrt{1-4y^2}\ge 0\Rightarrow \sqrt{1-4y^2}\ge -1\Rightarrow\; \text{True for all}\; y$

(2) Let me take the second condition :

$1 - \sqrt{1 - 4y^2} \ge 0\Rightarrow \sqrt{1 - 4y^2}\le 1\Rightarrow 1-4y^2\le 1\Rightarrow 4y^2\ge 0\Rightarrow y\in \mathbb{R}$.

$\small\texttt{I must be making mistake in step 1 or/and 2 above. But I can't find it.}$

 

[PeroK]

Yes, I said as much in my solution attempt in post #1. However, I am seeking to do this problem without the inspection of function $y$ itself, but finding out that $y\ge 0$ explicitly.



(1) Let me take the first condition :

$1+\sqrt{1-4y^2}\ge 0\Rightarrow \sqrt{1-4y^2}\ge -1\Rightarrow\; \text{True for all}\; y$

That implies $|y| \le \frac 1 2$.

(2) Let me take the second condition :

$1 - \sqrt{1 - 4y^2} \ge 0\Rightarrow \sqrt{1 - 4y^2}\le 1\Rightarrow 1-4y^2\le 1\Rightarrow 4y^2\ge 0\Rightarrow y\in \mathbb{R}$.

Likewise that implies the same condition on $y$.

$\small\texttt{I must be making mistake in step 1 or/and 2 above. But I can't find it.}$

There's no mistake. If you square $y$ and look for solutions, then you have lost the information that $y \ge 0$. For example:
$$y = 1 \Rightarrow y^2 = 1 \Leftrightarrow y = \pm 1$$In general, you may have to use all the available information. If you ignore some information (in this case that $y \ge 0$ from the definition of $y$), then that information may not necessarily be otherwise available.

 

[brotherbobby]

That implies |y|≤12.

I have found this (upper) limit using the method of determinants in my first post (#1).

Let me tell you what I am looking for, stating the problem and the answer.

Statement : Find the boundedness of the function $\quad\boldsymbol{f(x)=\dfrac{x^2}{x^4+1}}$

Answer : $\boxed{f(x)\in\left[0, \frac{1}{2}\right]}\quad\Large{\color{green}\checkmark}$

Doubt : The lower limit, viz. $f(x)\ge 0$ was found via inspection, which involves looking at the function and realising that it can be zero but never less. Is there an explicit way to determine this lower limit?

(Mind you, as my first post will reveal, the upper limit, viz. $\frac{1}{2}$, can be found explicitly)

 

[brotherbobby]

I hasten to add that I have found a way using the methods of calculus.

I copy and paste my working using $\;\texttt{myviewboard}^{\circledR}\;$, hoping am not violating anything. [I assume the limits are minima and maxima respectively, without finding the second derivative.]

Of course I had to use calculus. I'd been happier to have been able to find the lower limit, viz. 0, using the methods of algebra alone, same as I did for the upper limit 1/2.

 

[PeroK]

Doubt : The lower limit, viz. $f(x)\ge 0$ was found via inspection, which involves looking at the function and realising that it can be zero but never less. Is there an explicit way to determine this lower limit?

You could you prove that $x^2 \ge 0$ for all $x$ and $\frac 1 {1 + x^4} > 0$ for all $x$. And, prove that if $f(x) = \frac{g(x)}{h(x)}$ where $g(x) \ge 0$ and $h(x) > 0$, then $f(x) \ge 0$.

We only replace this proof by "inspection" because we've seen this result so many time already.

 

[pasmith]

I have found this (upper) limit using the method of determinants in my first post (#1).

Let me tell you what I am looking for, stating the problem and the answer.

Statement : Find the boundedness of the function $\quad\boldsymbol{f(x)=\dfrac{x^2}{x^4+1}}$

Answer : $\boxed{f(x)\in\left[0, \frac{1}{2}\right]}\quad\Large{\color{green}\checkmark}$

Doubt : The lower limit, viz. $f(x)\ge 0$ was found via inspection, which involves looking at the function and realising that it can be zero but never less. Is there an explicit way to determine this lower limit?

Does there exist $x \in \mathbb{R}$ such that $\frac{x^2}{1 + x^4} \leq 0$? Well, $1 + x^4 > 0$ for all $x \in \mathbb{R}$, so this can only happen when $x^2 \leq 0$.

 

[SammyS]

I have found this (upper) limit using the method of determinants in my first post (#1).

You used the discriminant, not the determinant.

 

[WWGD]

You also see $f(x)$ increases from $0$ to $1$ then decreases. Then use that $f(x)$ is even, so symmetric about $0$, and you can conclude minimum is at $x=0$, while maximum is at $x=1$.

 

[brotherbobby]

You used the discriminant, not the determinant.

Yes, pardon me.

 

[brotherbobby]

ou also see f(x) increases from 0 to 1 then decreases.

I would need the derivative of $f(x)$ wouldn't I to see that?

 

[WWGD]

I would need the derivative of $f(x)$ wouldn't I to see that?

Yes, indeed. My bad.