Question Number 71172 by aliesam last updated on 12/Oct/19
$${find}\:{fhe}\:{range} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{1}+\sqrt{{x}}} \\ $$
Commented by kaivan.ahmadi last updated on 12/Oct/19
![(√x)≥0⇒1+(√x)≥1⇒(4/(1+(√x)))≤4 but (4/(1+(√x)))>0 ⇒0<(4/(1+(√x)))≤4 ⇒R_f =(0,4]](https://www.tinkutara.com/question/Q71173.png)
$$\sqrt{{x}}\geqslant\mathrm{0}\Rightarrow\mathrm{1}+\sqrt{{x}}\geqslant\mathrm{1}\Rightarrow\frac{\mathrm{4}}{\mathrm{1}+\sqrt{{x}}}\leqslant\mathrm{4} \\ $$$${but}\:\:\frac{\mathrm{4}}{\mathrm{1}+\sqrt{{x}}}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}<\frac{\mathrm{4}}{\mathrm{1}+\sqrt{{x}}}\leqslant\mathrm{4} \\ $$$$\Rightarrow{R}_{{f}} =\left(\mathrm{0},\mathrm{4}\right] \\ $$
Commented by aliesam last updated on 12/Oct/19
$${thank}\:{you}\:{sir} \\ $$