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

Solution:

Given, f(x) = sin x + cos x

f(x) can be rewritten as

f(x) = \(\sqrt{2}[\frac{sin(x)}{\sqrt{2}}+\frac{cos(x)}{\sqrt{2}}]\) --- (1)

We know, \(\frac{1}{\sqrt{2}}=cos(45^{\circ})=sin(45^{\circ})\)

So equation (1) becomes,

f(x) = \(\sqrt{2}[cos(45^{\circ})sin(x)+sin(45^{\circ})cos(x)]\)

We know, sin (A+B) = sinAcosB + cosAsinB

Now, f(x) = \(\sqrt{2}[sin(45^{\circ}+x)]\) --- (2)

We know that a derivative with respect to dependent variables is positive for an increasing function.

Differentiating (2) on both sides with respect to x we get,

f’(x) = \(\sqrt{2}[cos(45^{\circ}+x)]\)

f’(x) > 0

\(\sqrt{2}[cos(45^{\circ}+x)]\) > 0

Where x lies in the interval (0, 2π)

We know that cos(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 (1) and (2) we have

x∈(0, π/4)∪(5π/4, 2π)

Therefore, the function is decreasing in x∈(π/4, 5π/4).

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

Summary:

Consider the function on the interval (0, 2π). f(x) = sin x + cos x. The open intervals on which the function is decreasing are x∈(π/4, 5π/4).