Question

Find the range range of the following function .A) $f\left(x\right)=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{..................\infty }}}$
b) $y=\frac{e^x-e^{-x}}{e^x+e^{-x}},x\ge 0$
c)$f\left(x\right)=(3x^2-4x+5)$

Answer 1

(i) $f\left(x\right)=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{.....\infty }}}$

Let $y=f\left(x\right)$

$\Rightarrow y=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{....\infty }}}$

$\Rightarrow y=\sqrt{(1-\cos x)y}$

$\Rightarrow y^2=(1-\cos x)y$

$\Rightarrow y=(1-\cos x)$

$\because y\in [2,0]$

(ii) $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}=\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}$

at $x=0$,

$y=\frac{e^o-1}{e^o+1}=0$

at $x=\infty$ ,

$y=\frac{1-\frac{1}{\infty }}{1+\frac{1}{\infty }}=\frac{1}{1}=1$

$\Rightarrow y\in [0,1]$

(iii) $f\left(x\right)=(3x^2-4x+5)$

$\because$ It is quadratic with $a>0$

$\Rightarrow$ max value= $\infty$

For min value,

Let $y=f\left(x\right)$

$\Rightarrow y=3x^2-4x+5$

$\Rightarrow \frac{dy}{dx}=0$

$\Rightarrow 2\times 3x-4=0$

$\Rightarrow 6x-4=0$

$\Rightarrow x=\frac{2}{3}$

$Y_{min}=3{\left(\frac{2}{3}\right)}^2-3\times \frac{2}{3}+5$

$=\frac{4}{3}-\frac{6}{3}+5$

$=\frac{-2}{3}-5$

$=\frac{13}{3}$

$\Rightarrow y\in [\frac{13}{3},\infty ]$