If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I have tried most successfully is using trig substitutions...
For example, if I substitute $x = \sin \phi$, this yields $\sin \phi + \cos \phi$
But what should I do next? Or there are any methods else to solve the problem?
 A: Substitute $$x = \sin \theta$$ The expression is $$x + \sqrt{-x^2 + 1} = \sin \theta + |\cos \theta|$$
We want to find the max. value of $\sin \theta + |\cos \theta|$. Now, two cases: $\cos \theta < 0$ and $\cos \theta \geq 0$.
$$\textbf{Case 1:} \cos \theta < 0$$
We need to find the max. value for $\sin \theta - \cos \theta$ for $\theta \in (\frac{\pi}{2}, \frac{3 \pi}{2})$
Note that $\sin \theta - \cos \theta \leq \sqrt{2}$ always.
In this range, this value is achievable at $\theta = \frac{3 \pi}{4} + 2\pi n, n \in \mathbb{Z}$
$$\textbf{Case 2:} \cos \theta \geq 0$$
We need to find the max. value for $\sin \theta + \cos \theta$ for $\theta \in [0, \frac{\pi}{2}] \cup [\frac{3 \pi}{2}, 2 \pi)$
Note that $\sin \theta + \cos \theta \leq \sqrt{2}$ always.
In this range, this value is achievable at $\theta = \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}$
$$\textbf{Thus, the max. value for} \sin \theta + |\cos \theta| \textbf{ is } \sqrt{2} \textbf{ achieved at } x = \frac{1}{\sqrt{2}}$$
NOTE $1$: Note that mentioning the values for $\theta$ for which the max. value occurs is important. Min-max problems are two step problems:
$1.$ Show that some expression is bounded.
$2.$ Show that the bound is achievable for the values we are concerned with.
As an exercise, try to find the min. value of the original expression. By following the same case-work and not following step-$2$, one would arrive at the incorrect conclusion that $-\sqrt{2}$ is the min. value for $\sin \theta + |\cos \theta|$.
NOTE $2$: We use the fact that $$-\sqrt{a^2 + b^2} \leq a \sin \theta + b \cos \theta \leq \sqrt{a^2 + b^2}$$
Why is it true?
Consider the polar co-ordinates of $(a, b)$. Let it be $(r, \phi)$ where $r = \sqrt{a^2 + b^2}$. $$a = r \cos \phi$$ $$b = r \sin \phi$$ Substitute, and we get $$a \sin \theta + b \cos \theta = r \sin(\theta + \phi) = \sqrt{a^2 + b^2} \sin(\theta + \phi)$$
A: We have
$$(x + \sqrt{1 - x^2})^2
+ (x - \sqrt{1 - x^2})^2
= 2x^2 + 2(1 - x^2) = 2 \tag{1}$$
which results in
$$x + \sqrt{1 - x^2} \le \sqrt 2.$$
(Note: Alternatively, we my use $(a + b)^2 \le 2(a^2 + b^2)$ to get
$(x + \sqrt{1 - x^2})^2 \le 2[x^2 + (1 - x^2)] = 2$.)
Also, when $x = 1/\sqrt 2$, we have $x + \sqrt{1 - x^2} = \sqrt 2$.
(Note: From (1), letting $(x - \sqrt{1 - x^2})^2 = 0$, we get $x = \sqrt{1 - x^2}$ or $x = 1/\sqrt 2$.)
Thus, the maximum of $x + \sqrt{1 - x^2}$ on $[-1,1]$ is $\sqrt 2$.
A: For every $x\in[-1,1]$ there exists a unique angle $\alpha\in[-\pi/2,\pi/2]$ such that $x=\sin\alpha$. This is usually denoted as $\alpha=\arcsin x$.
With this assumption, you have $\cos\alpha\ge0$ and so
$$
\sqrt{1-x^2}=\sqrt{\cos^2\alpha}=\cos\alpha
$$
Thus the expression to maximize is
$$
f(\alpha)=\sin\alpha+\cos\alpha,\quad \alpha\in[-\pi/2,\pi/2]
$$
Now it's known that
$$
f(\alpha)=\sqrt{2}\sin(\alpha+\pi/4)
$$
which is the same as maximizing
$$
g(\beta)=\sqrt{2}\sin\beta,\quad \beta\in[-\pi/4,3\pi/4]
$$
The sine is increasing over $[-\pi/4,\pi/2]$ and decreasing over $[\pi/2,3\pi/4]$, so the maximum is for $\beta=\pi/2$.
The maximum is $g(\pi/2)=\sqrt{2}$.
The minimum is either at $\beta=-\pi/4$ or at $\beta=3\pi/4$, but the former value is negative and the latter is positive, so the minimum is $g(-\pi/4)=-1$.

Just for confirmation, if $F(x)=x+\sqrt{1-x^2}$, we have
$$
F'(x)=1-\dfrac{x}{\sqrt{1-x^2}}
$$
which vanishes for $\sqrt{1-x^2}=x$, so for $x=1/\sqrt{2}$. We have
$$
F(-1)=-1,\quad F(1/\sqrt{2})=\sqrt{2},\quad F(1)=1
$$
so we also know that the minimum is $-1$, attained at $x=-1$.
