What is the maximum?

$\mathcal{C}(x)=x+\sqrt{1-x^2}~~, -1\leq x\leq 1$ Put $x=\sin \Theta$ ,

$\mathcal{C}(\theta)=\sin \Theta +\cos \Theta~~,-\pi/2\leq \Theta \leq \pi/2$

$(\small{\color{#3D99F6}{\because -\pi/2\leq \Theta \leq \pi/2\therefore \sqrt{1-\sin^2 \Theta}=|\cos \Theta|=\cos \Theta}})$

By Cauchy Shwarz or elementary trigonometry we know $\mathcal C(\Theta)\leq \sqrt 2$ .

Hence ,

$\large \mathcal{C}(\theta)_{\text{max}}=\sqrt2~~\small{\text{(For } \Theta=\pi/4)}$

For stationary points, $\frac{d}{dx}(x+\sqrt{1-x^2})=0$

$1-2x\cdot\frac{1}{2}(1-x^2)^{-\frac{1}{2}}=0$

$1=x(1-x^2)^{-\frac{1}{2}}$

$1=x^2(1-x^2)^{-1}$

$1-x^2=x^2\Rightarrow x=\pm\frac{1}{\sqrt{2}}$

Applying the second derivative test, we find that $x=\frac{1}{\sqrt{2}}$ gives a maximum value of $\boxed{\sqrt{2}}$ .

First I noticed that the maximum has to occur in the interval $0\leq x\leq 1$ . This allowed me to apply the Quadratic mean - Arithmetic mean inequality: $\frac{x+\sqrt{1-x^2}}{2}\leq\sqrt{\frac{x^2+1-x^2}{2}} \Rightarrow x+\sqrt{1-x^2}\leq \sqrt{2}$ . The maximum occurs when $x=\sqrt{1-x^2}$ or, equivalently, $x=1/\sqrt{2}$ . Therefore the maximum is the RHS of the inequality, $\sqrt{2}$ .