How do I find the max of $\sqrt{x+1}-\sqrt{x}$

Question

I know the answer is (0,1) ,and I tried setting the derivitave to zero but for some reason I get no solution.

Answer 1

The function $f(x) = \sqrt{x+1}-\sqrt{x}$ is defined for $x \ge 0$, and differentiable for $x > 0$. As it turns out, there is no point $x > 0$ at which $f'(x) = 0$ or $f'(x)$ is undefined, which is why you got no solution.

Since $f(x)$ is continuous for $x \ge 0$ and differentiable for $x > 0$, this means that either the maximum is obtained either at $x = 0$, or there is no maximum. To decide which, compute $f(0)$ and $\displaystyle\lim_{x \to \infty}f(x)$ and see which one is larger.

Alternatively, note that $f(x) = \sqrt{x+1}-\sqrt{x} = \dfrac{1}{\sqrt{x+1}+\sqrt{x}}$, where the denominator is strictly increasing for $x \ge 0$. What does this tell you about $f(x)$ and where its maximum is?

Answer 2

$$\frac{d}{dx}\sqrt{x+1}-\sqrt{x}=\frac1{2\sqrt{x+1}}-\frac{1}{2\sqrt x}=\frac{1}{2\sqrt{x+1}}\frac{2\sqrt x}{2\sqrt x}-\frac{1}{2\sqrt x}\frac{2\sqrt{x+1}}{2\sqrt{x+1}}$$ Can you take it from there?