The function $f(x)=\sin x+\cos x$ on $[0,2\pi]$ is decreasing on the i…

Question

The function $f(x)=\sin x+\cos x$ on $[0,2\pi]$ is decreasing on the interval

Accepted Answer

C. $\left(	frac{\pi}{4},	frac{5\pi}{4}ight)$

1. $f'(x)=\cos x-\sin x$.
2. $f'(x)=0$ when $	an x=1$, i.e. $x=	frac{\pi}{4},	frac{5\pi}{4}$.
3. On $\left(	frac{\pi}{4},	frac{5\pi}{4}ight)$, $\sin x>\cos x$, so $f'(x)<0$ (decreasing).
4. On the two outer pieces $f'(x)>0$, ruling out A and B.
_Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.10_

Answer

$\left[0,	frac{\pi}{4}ight)$ only

Answer

$\left(	frac{5\pi}{4},2\piight]$ only

Answer

$\left[0,2\pi\right]$ fully

The function $f(x)=\sin x+\cos x$ on $[0,2\pi]$ is decreasing on the interval

Keep practising Continuity, Differentiability & Derivatives