The range of $f(x)=\frac{1}{1+\sqrt{x}}$ is equal to
Answer & explanation
Correct answer: option 3
$0≤\sqrt{x}<∞⇒1≤1+\sqrt{x}<∞⇒0<\frac{1}{1+\sqrt{x}}≤1⇒0<f(x)≤1$
The range of $f(x)=\frac{1}{1+\sqrt{x}}$ is equal to
Correct answer: option 3
$0≤\sqrt{x}<∞⇒1≤1+\sqrt{x}<∞⇒0<\frac{1}{1+\sqrt{x}}≤1⇒0<f(x)≤1$