Question

Determine range of the function $y=4^x-2^x+1$
Range is (a,b)
then b-a=?

Answer 1

$y=4^x-2^x+1$
Above is defined for all $xϵR$
$\therefore t^2-t+1-y=0$ where $t=2^x$
$\therefore t=2^x=\frac{1}{2}(1\pm \sqrt{4y}-3)$
$\therefore x={\log }_2\left\{\frac{1\pm \sqrt{4y-3}}{2}\right\}$
Now x is real and hence we must have
$4y-3\ge 0$ and $1\pm \sqrt{4y-3}>0$ by def. of log x.
$\therefore y\ge \frac{3}{4}$ and $1-1\pm \sqrt{4y-3}>0$
$1+\sqrt{4y-3}$ is clearly > 0
$\therefore y\ge \frac{3}{4}$ and $\sqrt{4y-3}$ is clearly < 0
$\therefore y\ge \frac{3}{4}$ and $\sqrt{4y-3}<1$
or $\therefore 4y-3<1$ or $y<1$
$\therefore \le y<1$$\therefore Range=(\frac{3}{4},1)$