Q.Show that the function $f(x) = \tan^{-1}(\sin x + \cos x)$, $x>0$ is strictly increasing in the interval $\left(0, \dfrac{\pi}{4}\right)$.

Definitions. On an interval $I$, $f$ is increasing if $a<b\Rightarrow f(a)\le f(b)$ for all $a,b\in I$, and decreasing if $a<b\Rightarrow f(a)\ge f(b)$. ("Strictly" increasing/decreasing replaces $\le/\ge$ with $</>$.)

Theorem 7.7 (the working test). If $f$ is differentiable on the open interval $(a,b)$:

• $f'(x)\ge0$ for all $x\in(a,b)$ $\Rightarrow$ $f$ is increasing on $(a,b)$;
• $f'(x)>0$ for all $x\in(a,b)$ $\Rightarrow$ $f$ is strictly increasing on $(a,b)$;
• $f'(x)\le0$ (resp. $<0$) throughout $\Rightarrow$ $f$ is (strictly) decreasing.

A subtle converse warning. Strict monotonicity does not force $f'(x)>0$ everywhere — e.g. $f(x)=x^3$ is strictly increasing on all of $\mathbb{R}$, yet $f'(0)=0$. So the theorem's arrow only runs one way in general (a non-strict $f'\ge0$ that is zero only at isolated points is enough for strict monotonicity, as in this example).

Stationary vs. critical points.

• A stationary point $\big(x_0,f(x_0)\big)$ is where $f'(x_0)=0$.
• A critical point $\big(x_0,f(x_0)\big)$ is where $f'(x_0)=0$ or $f'(x_0)$ fails to exist.

Every stationary point is a critical point, but not conversely — e.g. $f(x)=|x-17|$ has a critical point at $x=17$ (the derivative doesn't exist there, a "corner"), which is not a stationary point.

Working procedure (as used throughout Exercise 7.6 Q2): …