# Determine the open intervals on which the graph of the function y = 3x-\tan 2x, (-\frac{\pi}{4},...

## Question:

Determine the open intervals on which the graph of the function {eq}y = 3x-\tan 2x, (-\frac{\pi}{4}, \frac{\pi}{4}) {/eq} is concave upward or concave downward.

A. concave upward {eq}\left (-\frac{\pi}{2}, 1 \right ) {/eq}; concave downward {eq}\left ( 1, \frac{\pi}{2} \right ) {/eq}

B. concave upward {eq}\left ( -\frac{\pi}{2},-\frac{\pi}{6} \right ), \left ( \frac{\pi}{6}, \frac{\pi}{2} \right ) {/eq}; concave downward {eq}\left ( -\frac{\pi}{6}, \frac{\pi}{6} \right ) {/eq}

C. concave upward {eq}\left ( -\frac{\pi}{4}, 0 \right ) {/eq}; concave downward {eq}\left (0, \frac{\pi}{4} \right ) {/eq}

D. concave upward {eq}\left ( 0, \frac{\pi}{4} \right ) {/eq}; concave downward {eq}\left (-\frac{\pi}{4}, 0\right ) {/eq}

E. concave upward {eq}\left (-\frac{\pi}{6}, \frac{\pi}{6}\right ) {/eq}; concave downward {eq}\left (-\frac{\pi}{2},-\frac{\pi}{6}\right ),\left (\frac{\pi}{6}, \frac{\pi}{2}\right ) {/eq}

A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is {eq}\theta = \frac{3\pi}{11} {/eq} radians. The distance (in meters) the ball bearing rolls in f seconds is {eq}s(t) = 4.9(sin \theta)t^2 {/eq}. Determine the value of {eq}s'(t) {/eq} after one second. Round numerical values in your answer to one decimal place.

A. {eq}s'(t) = 8.5 {/eq}

B. {eq}s'(t) = 6.4 {/eq}

C. {eq}s'(t) = 9.7 {/eq}

D. {eq}s'(t) = 5.7 {/eq}

E. {eq}s'(t) = 7.4 {/eq}

## Concavity

The concavity of the curve tells us if a certain interval is concave up or concave down. This is calculated by first getting the values that makes the second derivative zero or undefined. These values will become the endpoints of the intervals that we will look into.

## Answer and Explanation:

1.

Let's start by getting the second derivative of the given function.

{eq}\displaystyle y(x) = 3x - \tan(2x) \\ \displaystyle y'(x) = (1)3x^{1-1} - \sec^2(2x) \cdot 2 \\ \displaystyle y'(x) = 3x^{0} - 2\sec^2(2x) \\ \displaystyle y'(x) = 3(1) - 2\sec^2(2x) \\ \displaystyle y'(x) = 3 - 2\sec^2(2x) \\ \displaystyle y''(x) = 0 - (2)2\sec (2x) \cdot \sec(2x)\tan(2x) \cdot 2 \\ \displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) {/eq}

Equate this to zero and solve for {eq}x {/eq}.

{eq}\displaystyle 0 = -8\sec^2 (2x) \tan(2x) \\ \displaystyle \frac{0}{-8} = \sec^2 (2x) \tan(2x) \\ \displaystyle 0 = \sec^2 (2x) \tan(2x) \\ \displaystyle 0 = \sec^2 (2x),\;\; 0 = \tan(2x) \\ \displaystyle \sqrt{0 }= \sqrt{\sec^2 (2x)},\;\; 0 = \tan(2x) \\ \displaystyle 0 = \sec(2x),\;\; 0 = \tan(2x) \\ \displaystyle \textrm{arcsec}(0)= 2x,\;\; \arctan(0) = 2x \\ \displaystyle \text{No solution},\;\; \arctan(0) = 2x \\ \displaystyle \arctan(0) = 2x \\ \displaystyle 0 + \pi n = 2x \\ \displaystyle \pi n = 2x \\ \displaystyle \frac{\pi n}{2} = x {/eq}

The only solution found in the interval {eq}\displaystyle \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) {/eq} is at {eq}n = 0 {/eq}, hence the critical point of the function is

{eq}\displaystyle x = \frac{\pi (0)}{2} \\ \displaystyle x = 0 {/eq}

This will become the endpoint of the intervals that we will look into. The intervals are {eq}\displaystyle \left(-\frac{\pi}{4}, 0\right) \text{ and } \left(0, \frac{\pi}{4}\right) {/eq}. We will choose a value within these intervals and substitute them to the second derivative. If it results to a positive value, then the interval is concave up. If it results to a negative value, then the interval is concave down.

For the first interval {eq}\displaystyle \left(-\frac{\pi}{4}, 0\right) {/eq}, let's choose {eq}\displaystyle x = - \frac{\pi}{6} {/eq}:

{eq}\displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8\sec^2 (2\left ( -\frac{\pi}{6} \right )) \tan(2\left ( -\frac{\pi}{6} \right )) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(\sec \left ( -\frac{\pi}{3} \right ))^2 \tan\left ( -\frac{\pi}{3} \right ) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(2)^2 (-\sqrt{3}) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(4) (-\sqrt{3}) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = 32\sqrt{3} \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = 32\sqrt{3} > 0 \therefore \left(-\frac{\pi}{4},0 \right )\text{ is concave up} \\ {/eq}

For the last interval {eq}\displaystyle \left(0, \frac{\pi}{4}\right) {/eq}, let's choose {eq}\displaystyle x = \frac{\pi}{6} {/eq}:

{eq}\displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -8\sec^2 (2\left ( \frac{\pi}{6} \right )) \tan(2\left (\frac{\pi}{6} \right )) \\ \displaystyle y''\left (\frac{\pi}{6} \right ) = -8(\sec \left (\frac{\pi}{3} \right ))^2 \tan\left (\frac{\pi}{3} \right ) \\ \displaystyle y''\left (\frac{\pi}{6} \right ) = -8(2)^2 (\sqrt{3}) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -8(4) (\sqrt{3}) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -32\sqrt{3} \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -32\sqrt{3} < 0 \therefore \left(0, \frac{\pi}{4}\right )\text{ is concave down} {/eq}

The answer is C.

2.

Let's start by getting the first derivative of the function for distance.

{eq}\displaystyle s(t) = 4.9 (\sin \theta ) t^2 \\ \displaystyle s'(t) = (2)4.9 (\sin \theta ) t^{2-1} \\ \displaystyle s'(t) = 9.8 (\sin \theta ) t^{1} \\ \displaystyle s'(t) = 9.8 (\sin \theta ) t {/eq}

Then, substitute {eq}\displaystyle \theta = \frac{3\pi}{11} {/eq} and {eq}t = 1 {/eq}.

{eq}\displaystyle s'(1) = 9.8 (\sin \left ( \frac{3\pi}{11} \right ) ) (1) \\ \displaystyle s'(1) = 9.8(0.75574) \\ \displaystyle s'(1) = 7.40 {/eq}

The answer is E.