# Finding the maximum value without using derivatives Find the maximum value of f(x)=2 sqrt(x)- sqrt(x+1)- sqrt(x-1) without using derivatives.

Finding the maximum value without using derivatives
Find the maximum value of
$f\left(x\right)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$
without using derivatives.
The domain of $f\left(x\right)$ is $x\in \left[1,\mathrm{\infty }\right)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D\left(f\right)$ and the maximum value is $f\left(1\right)=2-\sqrt{2}$. However, this uses derivatives.
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Jazmin Bryan
Note that
$\begin{array}{rl}f\left(x\right)& =\left(\sqrt{x}-\sqrt{x-1}\right)-\left(\sqrt{x+1}-\sqrt{x}\right)\\ & =\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x+1}+\sqrt{x}}\\ & =\frac{\sqrt{x+1}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}\\ & =\frac{2}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\end{array}$
which is a decreasing function of x.

Andreasihf
$f\left(x\right)=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x+1}+\sqrt{x}}=$
$=\frac{2}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\le$
$=\frac{2}{\left(1+\sqrt{2}\right)\sqrt{2}}=2-\sqrt{2}.$
The equality occurs for $x=1$, which says that we got a maximal value.