# For the following function, find the interval(s) of increasing/decreasing, concavity, and all...

## Question:

For the following function, find the interval(s) of increasing/decreasing, concavity, and all local max/min using an appropriate test. Ignore critical points which are not local extreme.

{eq}f(x) = x + \sin\left(2x\right) \quad {/eq} over {eq}\;\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right] {/eq}

## Nature of Intervals

The nature of intervals can be known through differentiation. The first derivative is used to know if the interval is increasing or decreasing. While the second derivative is used to know the concavity of the interval. Using these information, local minimum or maximum points can be known.

## Answer and Explanation:

Let's begin by getting the critical points of the function in the given domain. These points will be used as the endpoints of the intervals that we will look into. The critical points are the points in the domain where the first derivative of the function is either zero or undefined. Let's get the first derivative of the function.

{eq}f(x) = x + \sin\left(2x\right) \\ \displaystyle f'(x) = 1 + \cos(2x) \cdot 2 \\ \displaystyle f'(x) = 1 + 2\cos(2x) {/eq}

Let's equate this to zero and solve for {eq}x {/eq}.

{eq}\displaystyle 0 = 1 + 2\cos(2x) \\ \displaystyle -1 = 2\cos(2x) \\ \displaystyle \frac{-1}{2} = \cos(2x) \\ \displaystyle 2x = \arccos\left (-\frac{1}{2} \right ) \\ \displaystyle 2x = \frac{2\pi}{3}+ 2\pi n, \frac{4\pi}{3} + 2\pi n \text{ where} n \in \mathbb{Z} \\ \displaystyle x = \frac{2\pi}{3(2)} + \pi n, \frac{4\pi}{3(2)} + \pi n \\ \displaystyle x = \frac{\pi}{3} + \pi n, \frac{2\pi}{3} + \pi n {/eq}

Since we are given a domain, the critical points are

{eq}\displaystyle x = \frac{\pi}{3}, -\frac{\pi}{3} {/eq}

These points are possible local minimum or maximum. We will also use them as the endpoints of the intervals to know whether the point is minimum or maximum. Since we have two points, there are three intervals namely {eq}\displaystyle \left [-\frac{\pi}{2}, -\frac{\pi}{3} \right ), \left (-\frac{\pi}{3}, \frac{\pi}{3} \right ) {/eq} and {eq}\displaystyle \left (\frac{\pi}{3}, \frac{\pi}{2} \right ] {/eq}. We will choose a value within these intervals and substitute to the first derivative. If it results to a positive number, then the interval is increasing. If it results to a negative number, then the interval is decreasing.

First interval {eq}\displaystyle \left [-\frac{\pi}{2}, -\frac{\pi}{3} \right ) {/eq}, let {eq}\displaystyle x =- \frac{\pi}{2} {/eq}:

{eq}\displaystyle f'(x) = 1 + 2\cos(2x) \\ \displaystyle f'\left ( -\frac{\pi}{2} \right ) = 1 + 2\cos(2\left ( -\frac{\pi}{2} \right )) \\ \displaystyle f'\left ( -\frac{\pi}{2} \right ) = 1 + 2\cos(-\pi) \\ \displaystyle f'\left ( -\frac{\pi}{2} \right ) = 1 + 2(-1) \\ \displaystyle f'\left ( -\frac{\pi}{2} \right ) = 1 - 2 \\ \displaystyle f'\left ( -\frac{\pi}{2} \right ) = -1 < 0 \therefore \text{ the interval is decreasing} \\ {/eq}

Second interval {eq}\displaystyle \left (-\frac{\pi}{3}, \frac{\pi}{3} \right ) {/eq}, let {eq}\displaystyle x = 0 {/eq}:

{eq}\displaystyle f'(x) = 1 + 2\cos(2x) \\ \displaystyle f'\left (0 \right ) = 1 + 2\cos(2\left ( 0 \right )) \\ \displaystyle f'\left (0 \right ) = 1 + 2\cos(0) \\ \displaystyle f'\left (0 \right ) = 1 + 2(1) \\ \displaystyle f'\left (0 \right ) = 1 + 2 \\ \displaystyle f'\left (0 \right ) = 3 > 0 \therefore \text{ the interval is increasing} \\ {/eq}

Last interval {eq}\displaystyle \left (\frac{\pi}{3}, \frac{\pi}{2} \right ] {/eq}, let {eq}\displaystyle x = \frac{\pi}{2} {/eq}:

{eq}\displaystyle f'(x) = 1 + 2\cos(2x) \\ \displaystyle f'\left ( \frac{\pi}{2} \right ) = 1 + 2\cos(2\left ( \frac{\pi}{2} \right )) \\ \displaystyle f'\left ( \frac{\pi}{2} \right ) = 1 + 2\cos(\pi) \\ \displaystyle f'\left ( \frac{\pi}{2} \right ) = 1 + 2(-1) \\ \displaystyle f'\left ( \frac{\pi}{2} \right ) = 1 - 2 \\ \displaystyle f'\left ( \frac{\pi}{2} \right ) = -1 < 0 \therefore \text{ the interval is decreasing} \\ {/eq}

Since the intervals on the sides of the points changes, the points are local min and max. Specifically:

{eq}\displaystyle \text{local min}: -\frac{\pi}{3} \\ \displaystyle \text{local max}: \frac{\pi}{3} \\ \displaystyle \text{decreasing}: \displaystyle \left [-\frac{\pi}{2}, -\frac{\pi}{3} \right ) \cup \left (\frac{\pi}{3}, \frac{\pi}{2} \right ] \\ \displaystyle \text{increasing}: \left (-\frac{\pi}{3}, \frac{\pi}{3} \right ) {/eq}

Now, to get the concavity and the intervals that we will look into, we will use second derivative. We will begin by getting the inflection points which are the values that make the second derivative zero or undefined.

{eq}\displaystyle f'(x) = 1 + 2\cos(2x) \\ \displaystyle f''(x) = 0 + 2(-\sin(2x))\cdot 2 \\ \displaystyle f''(x) = -4\sin(2x) \\ \displaystyle 0 = -4\sin(2x) \\ \displaystyle \frac{0}{-4} = \sin(2x) \\ \displaystyle 0 = \sin(2x) \\ \displaystyle 2x = \arcsin(0) \\ \displaystyle 2x = 0, \pi, -\pi \\ \displaystyle x = \frac{0}{2}, \frac{\pi}{2}, -\frac{\pi}{2} \\ \displaystyle x = 0, \frac{\pi}{2},-\frac{\pi}{2} {/eq}

The intervals that we will look into are {eq}\displaystyle \left (-\frac{\pi}{2},0 \right ) {/eq} and {eq}\displaystyle \left ( 0, \frac{\pi}{2} \right ) {/eq}. We will also choose a value within the interval and substitute it to the second derivative. If the result is a positive number, then the interval is concave up. If the result is a negative number, then the interval is concave down.

First interval {eq}\displaystyle \left (-\frac{\pi}{2},0 \right ) {/eq}, let {eq}\displaystyle x = -\frac{\pi}{6} {/eq}:

{eq}\displaystyle f''(x) = -4\sin(2x) \\ \displaystyle f''\left ( -\frac{\pi}{6} \right ) = -4\sin(2\left ( -\frac{\pi}{6} \right )) \\ \displaystyle f''\left ( -\frac{\pi}{6} \right ) = -4\sin\left(-\frac{\pi}{3} \right ) \\ \displaystyle f''\left ( -\frac{\pi}{6} \right ) = -4\left(-\frac{\sqrt{3}}{2} \right ) \\ \displaystyle f''\left ( -\frac{\pi}{6} \right ) = 2\sqrt{3} > 0 \therefore \text{ the interval is concave up} \\ {/eq}

Last interval {eq}\displaystyle \left (0, \frac{\pi}{2} \right ) {/eq}, let {eq}\displaystyle x = \frac{\pi}{6} {/eq}:

{eq}\displaystyle f''(x) = -4\sin(2x) \\ \displaystyle f''\left ( \frac{\pi}{6} \right ) = -4\sin(2\left ( \frac{\pi}{6} \right )) \\ \displaystyle f''\left ( \frac{\pi}{6} \right ) = -4\sin\left(\frac{\pi}{3} \right ) \\ \displaystyle f''\left ( \frac{\pi}{6} \right ) = -4\left(\frac{\sqrt{3}}{2} \right ) \\ \displaystyle f''\left ( \frac{\pi}{6} \right ) = -2\sqrt{3} < 0 \therefore \text{ the interval is concave down} {/eq}

In conclusion, intervals are:

{eq}\displaystyle \text{concave up}: \left (-\frac{\pi}{2},0 \right )\\ \displaystyle \text{concave down}: \left (0, \frac{\pi}{2} \right ) {/eq}