Question

Consider the function on the interval (0, 2π). f(x) = sin x + cos x (a) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.)

Answer 1

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 $\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)$

Using these result in equation a we get

f(x) = $\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)$

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) = $f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0$

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)