# Find the intervals in which the function f given by $f(x) =\sin x + \cos x, 0 \leq x \leq 2\pi$ is strictly increasing or strictly decreasing.

## 1 Answer

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
$f(x)=\sin x+\cos x$
$f'(x)=\cos x-\sin x$
When $f'(x)=0$
$\cos x-\sin x=0$
$\cos x=\sin x$
Given that $x=\large\frac{\pi}{4},\frac{5\pi}{4}$ as $0\leq x\leq 2\pi$
The points $x=\large\frac{\pi}{4}$ and $x=\large\frac{5\pi}{4}$ divides the interval $[0,2\pi]$ into three disjoint intervals.
(i.e) $[(0,\large\frac{\pi}{4}),(\frac{\pi}{4},\frac{5\pi}{4})$ and $(\large\frac{5\pi}{4}$$,2\pi)] Step 2: f'(x)>0 if x\in [(0,\large\frac{\pi}{4})$$\cup (\large\frac{5\pi}{4}$$,2\pi)] Or f is strictly increasing in the intervals [(0,\large\frac{\pi}{4}) and \large\frac{5\pi}{4}$$,2\pi]$
Also $f'(x)<0$ if $x\in (\large\frac{\pi}{4},\frac{5\pi}{4})$
$f$ is strictly decreasing in $(\large\frac{\pi}{4},\frac{5\pi}{4})$
answered Nov 15, 2013

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