# Determine the open intervals on which the graph is concave upward or concave downward. \\...

## Question:

Determine the open intervals on which the graph is concave upward or concave downward.

{eq}y=6x+\frac{2}{\sin x}, \ (-\pi, \pi){/eq}

## Differentiation:

Taking the derivative of a function involves finding the rate of change between two points. This is symbolized by {eq}\frac{d}{dx}f(x) {/eq}. If the derivative is positive, the function is increasing on an interval. If the derivative is negative, the function is decreasing on an interval. The second derivative of a function involves taking a derivative of the derivative. If the second derivative is positive, the function is concave up. If the second derivative is negative, the function is concave down.

## Answer and Explanation:

{eq}y = 6x+\frac{2}{\sin x}, (-\pi, \pi) {/eq}

The first step here is to find the derivative of {eq}y {/eq} in order to find the second derivative.

{eq}\begin{align*} y' &= \frac{d}{dx}(y = 6x+\frac{2}{\sin x}) \\ &= \frac{d}{dx}\left(6x+2\csc \left(x\right)\right) \\ &= 6\cdot \:1+2\left(-\cot \left(x\right)\csc \left(x\right)\right) \\ &= 6-2\cot \left(x\right)\csc \left(x\right) \\ \end{align*} {/eq}

Now the second derivative can be found and solved for possible critical points by setting the second derivative equal to {eq}0 {/eq}. If {eq}y'' > 0 {/eq}, {eq}y {/eq} is concave up. If {eq}y'' < 0 {/eq}, {eq}y {/eq} is concave down.

{eq}\begin{align*} y'' = \frac{d}{dx}(y' = 6-2\cot \left(x\right)\csc \left(x\right)) = 0 &\Rightarrow y'' = 0-2\left(\frac{d}{dx}\left(\cot \left(x\right)\right)\csc \left(x\right)+\frac{d}{dx}\left(\csc \left(x\right)\right)\cot \left(x\right)\right) = 0 &\Rightarrow y'' = 2\left(\left(-\csc ^2\left(x\right)\right)\csc \left(x\right)+\left(-\cot \left(x\right)\csc \left(x\right)\right)\cot \left(x\right)\right) = 0 &\Rightarrow y'' = 2\left(-\csc ^3\left(x\right)-\cot ^2\left(x\right)\csc \left(x\right)\right) = 0 \\ &\Rightarrow y'' = 2\left(-\csc ^3\left(x\right)-\cot ^2\left(x\right)\csc \left(x\right)\right) = 0 \\ &\Rightarrow y'' = 2(-\csc ^3\left(x\right)-\left(-1+\csc ^2\left(x\right)\right)\csc \left(x\right)) = 0 \\ &\Rightarrow y'' = 2(-\csc ^3\left(x\right)+\csc \left(x\right)-\csc ^3\left(x\right)) = 0 \\ &\Rightarrow y'' = 2(-\csc ^3\left(x\right)+\csc \left(x\right)-\csc ^3\left(x\right)) = 0 \\ &\Rightarrow y'' = 2\left(\csc \left(x\right)-2\csc ^3\left(x\right)\right) = 0 \\ &\Rightarrow \frac{1}{2}(2\left(\csc \left(x\right)-2\csc ^3\left(x\right)\right) = 0) \\ &\Rightarrow \csc \left(x\right)-2\csc ^3\left(x\right) = 0 \\ &\Rightarrow \csc(x)\left(1-2\csc ^2\left(x\right)\right) = 0 \\ &\Rightarrow \csc(x) = 0, 1-2\csc ^2\left(x\right) = 0 \\ &\Rightarrow \frac{1}{\sin(x)} = 0, 1-2\csc ^2\left(x\right)+2\csc^2(x) = 0+2\csc^2(x) \\ &\Rightarrow \frac{1}{\sin(x)} = 0 \text{ [ No possible solution]}, 1 = 2\csc^2(x) \\ &\Rightarrow \frac{1}{2}(1 = 2\csc^2(x)) \\ &\Rightarrow \frac{1}{2} = \csc^2(x) \\ &\Rightarrow \frac{1}{2} = \frac{1}{\sin^2(x)} \\ &\Rightarrow 2 = \sin^2(x) \\ &\Rightarrow \sqrt{2 = \sin^2(x)} \\ &\Rightarrow \sin(x) = \sqrt{2} \text{ [ Not in domain]} \\ \end{align*} {/eq}

The intervals under analysis for concavity are {eq}(-\pi, \pi) {/eq}.

For the interval {eq}(-\pi, \pi) {/eq}:

{eq}y'' = 2\left(\csc \left(\frac{\pi}{4}\right)-2\csc ^3\left(\frac{\pi}{4}\right)\right) \Rightarrow y'' = 2\left(1.414-2(2.828)\right) \Rightarrow y'' = 2(1.414-5.656) \Rightarrow y'' = 2(-4.242) \Rightarrow y'' = -8.484 {/eq}. Therefore, {eq}y'' {/eq} is concave down on the interval {eq}(-\pi, \pi) {/eq}.

#### Learn more about this topic:

from Math 104: Calculus

Chapter 10 / Lesson 6