- #1

brotherbobby

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- 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{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}##

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]##

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