## University Calculus: Early Transcendentals (3rd Edition)

$\mathrm{Range:}\ \ [2,3)$
$\mathrm{Remember:}\$ We can take square root only of the positive numbers, i.e $\ \ \sqrt{f(x)}\quad \Rightarrow \quad \:f(x)\ge 0$ To find the range of function we have to take a look at the given fraction $\ \ \frac{x^2}{x^2+4}.$ Since, by putting $\ \ x<0,$ the numerator and the denominator would yield a positive result. So the fraction as a whole is always positive. When we put $\ \ x=0,$ the value of fraction would be zero, as $\ \ \frac{0}{4}=0.$ As we increase the value of $\ x\$, the value of fraction would become closer and closer to 1, but it will never reach 1 because $\ \ x^2 < x^2+4\ \$ for any real number $\ x.$ So, we will have: $\mathrm{Minimum\: Value:}\ \ 2+0=2$ $\mathrm{Maximum\: Value:}\ \ \approx 2+1\approx 3\ \$ (the value will never reach 3). $\mathrm{Range:}\ \ [2,3)$