Question

Find the interval in which the function given by $f\left( x \right) = \sin x + \cos x;{\text{ 0}} \leqslant {\text{x}} \leqslant {\text{2}}\pi$ is
A. Increasing
B. Decreasing

Answer 1

In order to solve the above question, we have to first differentiate the function $f\left( x \right)$ and then equate it to $0$. After deriving the value of the intervals, we will find the value of $f'\left( x \right)$ in these intervals. If the value of $f'\left( x \right)\rangle 0$ then the function is increasing but if the value of $f'\left( x \right)\langle 0$ then the function is decreasing.

Formula used:
The function is decreasing if $f'\left( x \right)\langle 0$
The function is increasing if $f'\left( x \right)\rangle 0$
Also, remember the formulas for derivation of $\cos x$ and $\sin x$ .

Complete step by step solution:
We are given that $f\left( x \right) = \sin x + \cos x;{\text{ 0}} \leqslant {\text{x}} \leqslant {\text{2}}\pi$ ,
Now we will differentiate both the sides with respect to $x$ ,
$f'\left( x \right) = \cos x - \sin x$
Put $f'\left( x \right) = 0$ , we get,

$$\cos x - \sin x = 0 \\\ \Rightarrow \cos x = \sin x \\\ \Rightarrow \dfrac{{\sin x}}{{\cos x}} = 1 \\\$$

Now,

$$\tan x = 1 \\\ \Rightarrow \tan x = \tan \dfrac{\pi }{4} = \tan \dfrac{{5\pi }}{4} \\\$$

From this we derive that
$x = \dfrac{\pi }{4},\dfrac{{5\pi }}{4}$ as $0 \leqslant x \leqslant 2\pi$ .
The interval $\left[ {0,2\pi } \right]$ is divided by the points $x = \dfrac{\pi }{4},\dfrac{{5\pi }}{4}$ into three parts $\left[ {\left( {0,\dfrac{\pi }{4}} \right),\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right),\left( {\dfrac{{5\pi }}{4},2\pi } \right)} \right]$ .
Now we will check the value of $f'\left( x \right)$ in each of these intervals.
In the first interval $\left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{{5\pi }}{4},2\pi } \right)$ , $f'\left( x \right)\rangle 0$
Therefore, $f\left( x \right)$ is increasing in the first and third interval.
In the second interval $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$ , $f'\left( x \right)\langle 0$ .
Therefore, $f\left( x \right)$ is decreasing in the second interval.
Hence, from this we get that
$f\left( x \right)$ is increasing in the first and third interval $\left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{{5\pi }}{4},2\pi } \right)$ and $f\left( x \right)$ is decreasing in the second interval $\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)$ .

Note: To calculate the sums relating to increasing and decreasing functions we must remember the differentiation of trigonometric functions as the first step of solving these sums is calculating the derivative of the function. Also, remember that if $f'\left( x \right)\langle 0$ then the function is a decreasing function but if $f'\left( x \right)\rangle 0$ then the function is said to be an increasing function.