The function g is defined for x>0 with g(1)=2, g'(x)=sin(x+1/x), and g"(x)=(1-1/x^2)cos(x+1/x)....

Question:

The function {eq}g {/eq} is defined for {eq}x>0 {/eq} with {eq}g(1)=2, g'(x)=sin(x+ \frac {1}{x}), {/eq} and {eq}g"(x)=(1-\frac {1}{x^2}) cos(x+ \frac {1}{x}). {/eq}

a. Find all values of {eq}x {/eq} in the interval {eq}0.12 \leq x \leq1 {/eq} at which the graph of g has a horizontal tangent line.

b. On what subintervals of {eq}(0.12,1) {/eq}, if any, is the graph of {eq}g {/eq} concave down? Justify your answer.

c. Write an equation for the line tangent to the graph of {eq}g {/eq} at {eq}x=0.3. {/eq}

d. Does the line tangent to the graph of {eq}g {/eq} at {eq}x=0.3 {/eq} lie above or below the graph of {eq}g {/eq} for {eq}0.3 <x<1 {/eq}?

Concavity and Inflection Points

A function {eq}f(x) {/eq} is concave up when the second derivative {eq}f''(x) {/eq} is positive. Concave up curves resemble a cup or a smile. The graph of the function is concave down when {eq}f''(x)<0 {/eq}. Concave down curves resemble a hill or a frown. The points where a function changes from concave up to concave down or from concave down to concave up are called inflection points.

Answer and Explanation:

a) The function {eq}g(x) {/eq} will have a horizontal tangent line when {eq}g'(x) = 0 {/eq}, since the derivative gives the slope of the tangent line, and a horizontal line has slope zero. We have

{eq}\sin\left(x+ \frac {1}{x}\right) = 0\\ x+\frac{1}{x} = n\pi\\ x^2 + 1 = n\pi x\\ x^2 - n\pi x + 1 = 0\\ x = \dfrac{n\pi\pm\sqrt{n^2\pi^2 - 4}}{2} {/eq}

Trying integer values for {eq}n {/eq} we have {eq}x = \dfrac{\pi-\sqrt{\pi^2 - 4}}{2} \approx 0.359\\ x = \dfrac{2\pi-\sqrt{4\pi^2-4}}{2} \approx 0.163 {/eq} are the only two values satisfying {eq}f'(x) = 0 {/eq} in the interval {eq}0.12\leq x \leq 1 {/eq}


b. The function will be concave down when {eq}g''(x) <0 {/eq}

First, finding where {eq}g''(x) = 0 {/eq} on the interval, we have

{eq}\left(1-\frac{1}{x^2}\right)\cos\left(x+\frac{1}{x}\right) = 0\\ {/eq}

which results in two equations to solve on the interval {eq}[0.12,1] {/eq}. First

{eq}1-\frac{1}{x^2} = 0\\ 1=\frac{1}{x^2}\\ x^2 = 1\\ x=1 {/eq}. The second equation to solve gives

{eq}\cos\left(x+\frac{1}{x}\right) = 0\\ x+\frac{1}{x} = (2n+1)\frac{\pi}{2}\\ x^2 + 1 = \frac{(2n+1)\pi}{2}x\\ x^2 - \frac{(2n+1)\pi}{2}x + 1 =0\\ x=\dfrac{\frac{(2n+1)\pi}{2}\pm\sqrt{\frac{(2n+1)^2\pi^2}{4} - 4}}{2} {/eq}

Substituting values for {eq}n {/eq} we have

{eq}x=\dfrac{\frac{3\pi}{2} - \sqrt{\frac{9\pi^2}{4} - 4}}{2} \approx 0.223\\ x=\dfrac{\frac{5\pi}{2} - \sqrt{\frac{25\pi^2}{4} - 4}}{2} \approx 0.129\\ {/eq}

Checking values on both sides of each possible inflection point, we have

{eq}g''(0.125) = \left(1-\frac{1}{0.125^2}\right)\cos\left(0.125+\frac{1}{0.125}\right)>0\\ g''(0.2) = \left(1-\frac{1}{0.2^2}\right)\cos\left(0.2+\frac{1}{0.2}\right)<0\\ g''(0.5) = \left(1-\frac{1}{0.5^2}\right)\cos\left(0.5+\frac{1}{0.5}\right)>0\\ {/eq}

So the function is concave down on the interval {eq}\left(\dfrac{\frac{5\pi}{2} - \sqrt{\frac{25\pi^2}{4} - 4}}{2} , \dfrac{\frac{3\pi}{2} - \sqrt{\frac{9\pi^2}{4} - 4}}{2}\right) {/eq} which is approximately {eq}(0.129, 0.223) {/eq}


c. To find the equation for the tangent line to the graph when {eq}x=0.3 {/eq} we need both {eq}g'(0.3) {/eq} and {eq}g(0.3) {/eq}

The function is given by {eq}g(x) = \int \sin\left(x+\frac{1}{x}\right) dx {/eq} however, this integral is not able to be expressed with elementary functions. We can however find {eq}g'(0.3) = \sin\left(0.3+ \frac{1}{0.3}\right) = \sin\left(\frac{109}{30}\right)\approx -0.472 {/eq}

So the equation of the tangent line is given by {eq}y - g(3) = \sin\left(\frac{109}{30}\right)(x-0.3) {/eq}


d. The tangent line to the graph of {eq}g(x) {/eq} at {eq}x=0.3 {/eq} is below the graph of the function for {eq}0.3<x<1 {/eq} because the function is concave up during that interval.


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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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