Answer:Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)
Step-by-step explanation:
given f(x) = sin(x) + cos(x)
f(x) can be rewritten as ![\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%20%5B%5Cfrac%7Bsin%28x%29%7D%7B%5Csqrt%7B2%7D%20%7D%2B%5Cfrac%7Bcos%28x%29%7D%7B%5Csqrt%7B2%7D%20%7D%20%20%5D..................%28a%29%5C%5C%5C%5C%5C%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%20%3D%20cos%2845%29%20%3D%20sin%2845%29%5C%5C%5C%5C)
Using these result in equation a we get
f(x) = ![\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%20%5B%20cos%2845%29sin%28x%29%2Bsin%2845%29cos%28x%29%5D%5C%5C%5C%5C%3D%20%5Csqrt%7B2%7D%20%5Bsin%2845%2Bx%29%5D..........%28b%29)
Now we know that for derivative with respect to dependent variable is positive for an increasing function
Differentiating b on both sides with respect to x we get
f '(x) = ![f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0](https://tex.z-dn.net/?f=f%20%27%28x%29%3D%5Csqrt%7B2%7D%20%20%5Cfrac%7Bdsin%2845%2Bx%29%7D%7Bdx%7D%5C%5C%20%5C%5Cf%27%28x%29%3D%5Csqrt%7B2%7D%20cos%2845%2Bx%29%5C%5C%5C%5Cf%27%28x%29%3E0%3D%3E%5Csqrt%7B2%7D%20cos%2845%2Bx%29%3E0)
where x∈(0,2π)
we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]
Thus for cos(π/4+x)>0 we should have
1) π/4 + x < π/2 => x<π/4 => x∈[0,π/4]
2) π/4 + x > 3π/2 => x > 5π/4 => x∈[5π/4,2π]
from conditions 1 and 2 we have x∈(0,π/4)∪(5π/4,2π)
Thus the function is decreasing in x∈(π/4,5π/4)