Question

Let $f$ be a continuous function on the closed, finite interval $[a,b]$ then
(i) $f$ is bounded on $[a,b]$ and (ii) $f$ attains both its maximum value $M$ and its minimum value $m$ on $[a,b]$ i.e. there is $c$, $dϵ[a,b]$ such that $f\left(c\right)=M,f\left(d\right)=m$.

Which of the following functions are not bounded

• A. $f\left(x\right)=\frac{2x}{1+x^2},[-2,2]$
• B. $f\left(x\right)=\frac{1-\cos x}{x^2},[-2,2]$
• C. $f\left(x\right)=\frac{x^2-8x+6}{4x+1},[0,5]$
• D. none of these

Answer 1

Answer: (D)

none of these

To show that a function is bounded in a particular closed interval we have to prove it is continuous on that interval.

A. Let $p\left(x\right)=2x$

$q\left(x\right)=1+x^2$ & $q\left(x\right)\ge 1\Rightarrow q\left(x\right)\ne 0$

We know that $p\left(x\right)$ & $q\left(x\right)$ are continious

Since they are polynomials

$\Rightarrow f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ is also continuous (Algebra of continuous functions)

B. At $x=0$

${lim}_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$

$1-\cos x$ & $x^2$ are continuous functions

$\Rightarrow \frac{1-\cos x}{x^2}$ is also continuous (ACF)

$f\left(x\right)=\frac{1-\cos x}{x^2}=2\frac{{\sin }^{2}x/2}{x^2}=\frac{{\sin }^{2}x/2}{2{\left(x/2\right)}^2}$

C. Let $p\left(x\right)=x^2-8x+6$

$q\left(x\right)=4x+1$

$4x+1\ge 1$ in $xϵ[0,5]$

$\Rightarrow q\left(x\right)\ne 0$

We know that $p\left(x\right)$ & $q\left(x\right)$ are continuous because they are polynomials

$\Rightarrow f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=\frac{x^2-8x+6}{4x+1}$ is also (A of CF)

Continuous.