## Exercises3.5Linear Approximation and Newton's Method

Solve the following problems.

###### 1.

The circumference of a sphere is measured to be 24 cm, with a possible error of 0.25 cm. Use the differential $dV$ to estimate the maximum error in the calculated volume $V\text{.}$

Hint
Observe that $\ds dr=\frac{1}{8\pi}\text{.}$
$\ds dV=\frac{72}{\pi^2}\text{.}$
###### 2.

The area $A$ of a square of side length $s$ is $A=s^2\text{.}$ Suppose $s$ increases by an amount $\Delta s=ds\text{.}$

1. Draw a square and then illustrate the quantity $dA$ on your diagram.

2. If $dA$ is used to approximate $\Delta A\text{,}$ illustrate the error of approximation on the same diagram.

Observe that $\Delta A=A(s+\Delta s)-A(s)=2s\Delta s+(\Delta s)^2$ and $dA=2s\Delta s\text{.}$

###### 3.

Let $f(x)=\sqrt{(x+4)^3}\text{.}$

1. Find the linear approximation to the function $f(x)=\sqrt{(x+4)^3}$ at $a=0\text{.}$

2. Use this approximation to estimate the number $\sqrt{(3.95)^3}\text{.}$ Is your estimate an overestimate or an underestimate?

Hint
What is the concavity of the function $f(x)\text{?}$
1. $L(x)=8+3x\text{.}$

2. $\sqrt{(3.95)^3} \approx 7.85\text{.}$ Underestimate.

Solution
1. Note that $f(0)=8\text{.}$ From $\ds f'(x)=\frac{3}{2}\sqrt{x+4}$ it folows that $f'(0)=3\text{.}$ Thus the linearization of $f$ at $a=0$ is $L(x)=8+3x\text{.}$

2. For $x$ “close” to $0$ we have that $\ds f(x)=\frac{3}{2}\sqrt{(x+4)^3}\approx L(x)\text{.}$ Thus $\sqrt{(3.95)^3}=f(-0.05)\approx L(-0.05)=8-0.15=7.85\text{.}$ Since $\ds f"(x)=\frac{3}{4\sqrt{x+4}}>0$ we conclude that, in the neighbourhood of $x=0\text{,}$ the graph of the function $f$ is above the tangent line at $x=0\text{.}$ Thus $L(-0.05)$ is an underestimate.

###### 4.
1. Use linear approximation to estimate $\sqrt[3]{65}\text{.}$

2. Use concavity to state if your estimate in (a) is greater than or less than the exact value of $\sqrt[3]{65}\text{.}$ Explain.

1. $\ds \sqrt[3]{65}\approx L(65)=\frac{193}{48}\text{.}$

2. Overestimate.

###### 5.

Use linear approximation to estimate the value of $\sqrt[3]{26^2}\text{.}$ Express your answer a single fraction (for example, $\ds \frac{16}{729}$).

$\ds \sqrt[3]{26^2} \approx \frac{79}{9}\text{.}$
Solution

Let $\ds f(x)=x^{\frac{2}{3}}\text{.}$ Then $\ds f(x)=\frac{2}{3}x^{-\frac{1}{3}}\text{,}$ $f(27)=9\text{,}$ and $\ds f'(27)=\frac{2}{9}\text{.}$ Hence the linearization of the function $f$ at $a=27$ is $\ds L(x)=9+\frac{2}{9}(x-27)\text{.}$ It follows that $\ds \sqrt[3]{26^2}=f(26)\approx L(26)=9-\frac{2}{9}=\frac{79}{9}\text{.}$ (Note: MAPLE gives $\ds \frac{79}{9}\approx 8.777777778$ and $\sqrt[3]{26^2}\approx 8.776382955\text{.}$)

###### 6.

Use the linear approximation to approximate $(63)^{2/3}\text{.}$ Then use differentials to estimate the error.

$\ds (63)^{2/3}\approx\frac{95}{6}\text{.}$
Solution

Let $\ds f(x)=x^{\frac{2}{3}}\text{.}$ Then $\ds f(x)=\frac{2}{3}x^{-\frac{1}{3}}\text{,}$ $f(64)=16\text{,}$ and $\ds f'(64)=\frac{1}{6}\text{.}$ Hence the linearization of the function $f$ at $a=64$ is $\ds L(x)=16+\frac{1}{6}(x-64)\text{.}$ It follows that $\ds (63)^{2/3} =f(63)\approx L(63)=16-\frac{1}{6}=\frac{95}{6}\text{.}$ The error is close to the absolute value of the differential $\ds |dy|=|f'(64)\Delta x| = \frac{1}{6}\text{.}$ (Note: MAPLE gives $\ds \frac{95}{6}\approx 15.83333333$ and $\sqrt[3]{63^2}\approx 15.83289626\text{.}$)

###### 7.

Use linear approximation to estimate the value of $\sqrt{80}\text{.}$

$\ds \sqrt{80}\approx\frac{161}{18}\text{.}$
Solution

Let $\ds f(x)=\sqrt{x}\text{.}$ Then $\ds f(x)=\frac{1}{2\sqrt{x}}\text{,}$ $f(81)=9\text{,}$ and $\ds f'(81)=\frac{1}{18}\text{.}$ Hence the linearization of the function $f$ at $a=81$ is $\ds L(x)=9+\frac{1}{18}(x-81)\text{.}$ It follows that $\ds \sqrt{80}=f(80)\approx L(80)=9-\frac{1}{18}=\frac{161}{18}\text{.}$ (Note: MAPLE gives $\ds \frac{161}{18}\approx 8.944444444$ and $\sqrt{80}\approx 8.944271910\text{.}$)

###### 8.

Assume that $f$ is function such that $f(5)=2$ and $f^\prime(5)=4\text{.}$ Using a linear approximation to $f$ near $x=5\text{,}$ find an approximation to $f(4.9)\text{.}$

The linearization of the function $f$ at $a=5$ is $L(x)=2+4(x-5)\text{.}$ Thus $f(4.9)\approx L(4.9)=2-0.4=1.6\text{.}$

###### 9.

Suppose that we don't have a formula for $g(x)$ but we know that $g(2)=-4$ and $g^\prime (x)=\sqrt{x^2+5}$ for all $x\text{.}$

1. Use linear approximation to estimate $g(2.05)\text{.}$

2. Is your estimate in part (a) larger or smaller than the actual value? Explain.

1. $-3.85\text{.}$

2. Larger.

Solution
1. The linearization of the function $g$ at $a=2$ is $L(x)=-4+3(x-2)\text{.}$ Thus $g(2.05)\approx L(2.05)=-3.85\text{.}$

2. From $\ds g"(2)=\frac{2}{3}>0$ we conclude that the function $g$ is concave downward at $a=2\text{,}$ i.e. the graph of the function lies below the tangent line. Thus, the estimate is larger than the actual value.

###### 10.

Let $f(x)=\sqrt{1-x}\text{.}$

1. Find a linear approximation for the function $f(x)=\sqrt{1-x}$ valid for $x$ close to $0\text{.}$

2. Use your answer to find an approximate value for $\sqrt{0.9}\text{.}$

3. Find the tangent line to the graph of $f(x)=\sqrt{1-x}$ at $x=0\text{.}$

4. Sketch a graph to illustrate the relationship between $f(x)=\sqrt{1-x}$ and its linear approximation near $x=0\text{.}$

1. $\ds L(x)=1-\frac{x}{2}\text{.}$

2. $\ds \sqrt{0.9}\approx 1-\frac{1}{20}=\frac{19}{20}\text{.}$

3. $\ds y=-\frac{x}{2}+1\text{.}$

###### 11.

Find the linear approximation of the function $f(x)=\sqrt{1+x}$ at $a=3\text{,}$ and use it to estimate the value of $\sqrt{5}\text{.}$ Use a sketch to explain if this is an overestimate or underestimate of the actual value.

$\ds \sqrt{5}\approx L(4) = 2.25\text{.}$ Overestimate.
###### 12.

Let $f(x)=\sqrt{1+2x}\text{.}$

1. Find the linear approximation of $f(x)$ at $x=0\text{.}$

2. Use your answer to estimate the value of $\sqrt{1.1}\text{.}$

3. Is your estimate an over- or under-estimate?

1. $L(x)=1+x\text{.}$

2. $\ds \sqrt{1.1}=f(0.05)\approx L(0.05)=1.05\text{.}$

3. An over-estimate since $f$ is concave-down. MAPLE gives $\sqrt{1.1} \approx 1.048808848\text{.}$

###### 13.

Let $f(x)=\sqrt[3]{x+8}\text{.}$

1. Find a linear approximation to the function $f(x)=\sqrt[3]{x+8}$ at $a=0\text{.}$

2. Use this approximation to estimate the numbers $\sqrt[3]{7.95}$ and $\sqrt[3]{8.1}\text{.}$

1. $\ds L(x)=2+\frac{x}{12}\text{.}$

2. $\ds \sqrt[3]{7.95}\approx L(-0.05)=2-\frac{1}{240}=\frac{479}{240}$ and $\ds \sqrt[3]{8.1}\approx L(0.1)=2+\frac{1}{120}=\frac{243}{120}\text{.}$ (Note: MAPLE gives $\ds \frac{479}{240}\approx 1.995833333$ and $\sqrt[3]{7.95}\approx 1.995824623\text{.}$ Also, $\ds \frac{243}{120}\approx 2.025000000$ and $\sqrt[3]{8.1}\approx 2.008298850\text{.}$)

###### 14.

Let $f(x)=(1+x)^{100}\text{.}$

1. Construct the linear approximation to $f(x)=(1+x)^{100}\text{.}$

2. Use your approximation from (a) to estimate $(1.0003)^{100}\text{.}$

3. Is your estimate from (b) higher or lower than the true value? Explain.

1. For $a\in \mathbb{R}\text{,}$ $L(x) = (1+a)^{100} + 100(1+a)^{99}(x-a)\text{.}$

2. $(1.0003)^{100} \approx L(0.0003) = 1.03\text{.}$

3. Underestimate.

###### 15.

Let $f(x)=\sqrt[3]{27+3x}\text{.}$

1. Find the equation of the tangent line to the graph of the function $f(x)=\sqrt[3]{27+3x}$ at $x=0\text{.}$

2. Use your answer to estimate a value of $\sqrt[3]{30}\text{.}$

3. Draw a sketch to show how the graph of $f$ and its tangent line behave around the point where $x=0$ and the value of $x$ where the value in part (b) is obtained.

1. $\ds y=\frac{x}{9}+3\text{.}$

2. $\ds \sqrt[3]{30}\approx \frac{1}{9}+3=\frac{28}{9}\text{.}$ (Note: MAPLE gives $\ds \frac{28}{9}\approx 3.111111111$ and $\sqrt[3]{30}\approx 3.107232506\text{.}$)

###### 16.

Use linear approximation to estimate the value of $\ln 0.9\text{.}$

$\ln 0.9\approx -0.1\text{.}$
Solution

The linearization of the function $f(x)=\ln x$ at $a=1$ is given by $L(x)=x-1\text{.}$ Thus $\ln 0.9\approx L(0.9)=-0.1\text{.}$ (Note: MAPLE gives $\ln 0.9\approx -.1053605157\text{.}$)

###### 17.

Use a linear approximation to estimate the value of $\ds e^{-0.015}\text{.}$ Is your estimate too large or too small?

Hint
Take $f(x)=e^x\text{.}$

$e^{-0.015} \approx 0.985\text{.}$ Underestimate.

###### 18.

Let $f(x)=\ln x\text{.}$

1. Write the linear approximation for $f(x)=\ln x$ around 1.

2. Compute the approximated value for $\exp (-0.1)$ using linear approximation.

1. $L(x)=x-1\text{.}$

2. Let $x=\exp (-0.1)\text{.}$ Then $\ln x=-0.1\approx L(x)=x-1\text{.}$ Thus $x\approx 0.9\text{.}$ (Note: MAPLE gives $\exp (-0.1)\approx 0.9048374180\text{.}$)

###### 19.

Using the function $f(x)=x^{1/3}$ and the technique of linear approximation, give an estimate for $1001^{1/3}\text{.}$

Solution

$\ds L(x)=10+\frac{1}{300}(x-1000)$ implies $\ds 1001^{1/3}\approx L(1001)=\frac{3001}{300}\text{.}$ (Note: MAPLE gives $\ds \frac{3001}{300}\approx 10.00333333$ and $\sqrt[3]{1001}\approx 10.00333222\text{.}$)

###### 20.

Let $\ds f(x)=\sqrt{x}+\sqrt[5]{x}\text{.}$

1. Use linear approximation to determine which of the following is nearest the value of $f(1.001)\text{:}$

\begin{equation*} \begin{array}{lllll} 2.0001\amp 2.0002\amp 2.0003\amp 2.0005\amp 2.0007\\ 2.001\amp 2.002\amp 2.003\amp 2.005\amp 2.007 \end{array} \end{equation*}
2. At $x=1\text{,}$ is $f(x)$ concave up or concave down?

3. Based on your answer above, is your estimate of $f(1.001)$ too high or too low?

Solution
1. The linearization of the function $\ds f(x)=\sqrt{x}+\sqrt[5]{x}$ at $a=1$ is given by $\ds L(x)=2+\frac{7}{10}(x-1)\text{.}$ Thus $f(1.001)\approx L(1.001)=2+0.7\cdot 0.001=2.0007\text{.}$

2. Note that the domain of the function $f$ is the interval $[0,\infty )\text{.}$ From $\ds f"(x) =-\frac{1}{4}x^{-\frac{3}{2}}-\frac{4}{25}x^{-\frac{9}{5}}$ it follows that $f$ is concave downwards on the interval $(0,\infty)\text{.}$

3. The graph of the function is below the tangent line at $a=1\text{,}$ so the estimate $f(1.001)\approx 2.0007$ is too high.

###### 21.
1. Find the linear approximation of $f(x)=\sin x$ about the point $x=\pi /6\text{.}$

2. Explain why $f$ satisfies the conditions of the Mean Value Theorem. Use the theorem to prove that $\ds \sin x\leq \frac{1}{2}+\left(x-\frac{\pi }{6}\right)$ on the interval $\ds \left[\frac{\pi }{6},x\right]$ where $\ds x>\frac{\pi }{6}\text{.}$

3. Is the differential $df$ larger or smaller than $\Delta f$ on the interval $\ds \left[\frac{\pi }{6},x\right]\text{?}$ Do not perform any calculations. Use only the results in part (a) and (b) to explain your answer.

1. $\ds L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi }{6}\right)\text{.}$

2. By the Mean Value Theorem, for $\ds x>\frac{\pi }{6}$ and some $\ds c\in \left(\frac{\pi }{6},x\right)\text{,}$ $\ds \frac{f(x)-f(\frac{\pi }{6})}{x-\frac{\pi }{6}}=\frac{\sin x-\frac{1}{2}}{x-\frac{\pi }{6}}=f^\prime(c)=\cos c\leq 1\text{.}$ Since $\ds x-\frac{\pi }{6}>0\text{,}$ the inequality follows.

3. From (a) and (b) it follows that, for $\ds x>\frac{\pi }{6}\text{,}$ $\ds \sin x\leq \frac{1}{2}+(x-\frac{\pi }{6})\lt \frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})=L(x)\text{.}$ Next, $\Delta f=f(x)-f(\frac{\pi }{6})=\sin x-\frac{1}{2}\lt L(x)-\frac{1}{2}=\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})= f^\prime (\frac{\pi }{6})\Delta x=df\text{.}$

###### 22.

Suppose that the only information we have about a function $f$ is that $f(1)=5$ and that the graph of its derivative is as shown in the Figure.

1. Use a linear approximation to estimate $f(0.9)$ and $f(1.1)\text{.}$

2. Are your estimates in part (a) too large or too small?

1. $f(0.9) \approx L(0.9) = 5.2\text{,}$ $f(1.1) \approx L(1.1) = 4.8\text{.}$

2. Too large.

###### 23.

Suppose that the only information we have about a function $f$ is that $f(1)=3$ and that the graph of its derivative is as shown in the Figure.

1. Use a linear approximation to estimate $f(0.9)$ and $f(1.1)\text{.}$

2. Are your estimates in part (a) too large or too small?

1. $f(0.9) \approx L(0.9) = 2.8\text{,}$ $f(1.1) \approx L(1.1) = 3.2\text{.}$

2. Too large.

###### 24.
1. State Newton's iterative formula that produces a sequence of approximations $x_1,x_2, x_3, \ldots$ to a root of function $f(x)\text{.}$

2. Find the positive root of the equation $\cos x=x^2$ using Newton's method, correct to 3 decimal points, with the first approximation $x_1=1\text{.}$

Solution

Let $f(x)=\cos x-x^2\text{.}$ Then $f'(x)=-\sin x-2x\text{.}$ Thus $\ds x_2=1-\frac{\cos 1-1}{-\sin 1-2}\approx 0.8382184099\text{,}$ $\ds x_2=0.8382184099-\frac{\cos 0.8382184099-0.8382184099^2}{-\sin 0.8382184099-2\cdot 0.8382184099}\approx \\ 0.8242418682\text{,}$ and $\ds x_3=0.8242418682-\frac{\cos 0.8242418682-0.8242418682^2}{-\sin 0.8242418682-2\cdot 0.8242418682}\approx \\ 0.8241323190\text{.}$ (Note: MAPLE gives $\cos 0.8241323190-0.8241323190^2\approx -1.59\cdot 10^{-8}\text{.}$)

###### 25.
1. State Newton's iterative formula that produces a sequence of approximations $x_0,x_1,x_2, \ldots$ to a solution of $f(x)=0\text{,}$ assuming that $x_0$ is given.

2. Draw a labeled diagram showing an example of a function $f(x)$ for which Newton's iterative formula fails to find a solution of $f(x)=0\text{.}$ Mark on your diagram $x_0\text{,}$ $x_1\text{,}$ and $x_2\text{.}$

Solution

Take $f(x)=\sqrt[3]{x}\text{,}$ $x_0=1\text{,}$ $x_1=-2\text{,}$ $x_2=4\text{,}$ and $x_3=-8\text{.}$

###### 26.
1. Explain how you can use Newton's Method to approximate the value of $\sqrt{5}\text{.}$

2. Explain which of the following choices is the best initial approximation when using Newton's Method as in (a):$-1\text{,}$ $0\text{,}$ or $1\text{?}$

3. Find the fourth approximation $x_4$ to the value of $\sqrt{5}$ using Newton's Method with the initial approximation $x_1$ you chose in (b).

1. We use Newton's Method to solve the equation $x^2-5=0\text{,}$ $x>0\text{.}$ From $f(x)=x^2-5$ and $f^\prime (x)=2x\text{,}$ Newton's Method gives $\ds x_{n+1}=x_n-\frac{x_n^2-5}{2x_n}=\frac{1}{2}\left( x_n+\frac{5}{x_n}\right)\text{.}$

2. A rough estimate of $\sqrt{5}$ gives a value that is a bit bigger than 2. Thus, take $x_1=1\text{.}$

3. $\ds x_2=3\text{,}$ $\ds x_3=\frac{7}{3}\text{,}$ $x_4=\frac{47}{21}\approx 2.23809\text{.}$ (Note: MAPLE gives $\sqrt{5}\approx 2.23606\text{.}$)

###### 27.

Apply Newton's method to $\ds f(x)=x^{1/3}$ with $x_0=1$ and calculate $x_1,x_2,x_3, x_4\text{.}$ Find a formula for $|x_n|\text{.}$ What happens to $|x_n|$ as $n\to \infty\text{?}$ Draw a picture that shows what is going on.

Solution

Let $\ds f(x)=x^{\frac{1}{3}}\text{.}$ Then Newton's method gives $\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f\,'(x_n)}= x_n-\frac{x_n^{\frac{1}{3}}}{\frac{1}{3}x_n^{-\frac{2}{3}}} = -2x_n\text{.}$ So $|x_{n+1}|=2|x_n|\text{.}$ This implies that if $x_0\neq0,\ |x_n| = 2^n|x_0|\rightarrow\infty$ as $n\rightarrow\infty\text{;}$ Newton's Method does not work in this case!

###### 28.
1. Find the Newton's method iteration formula to estimate $\sqrt[3]{68}\text{.}$

2. Provide an initial guess. Then explain, whether your initial guess is an over- or underestimate after the first iteration.

1. $\ds x_{n+1}=x_n-\frac{x_n^3-68}{3x_n^2}\text{.}$

2. $x_0=4\text{.}$ Underestimate.

###### 29.
1. Use linear approximation to estimate $\sqrt[3]{26}\text{.}$

2. The value of $\sqrt[3]{26}$ is approximately $x_1=3\text{.}$ Use Newton's method to find a better approximation, $x_2\text{,}$ to $\sqrt[3]{26}\text{.}$

1. $\sqrt[3]{26}\approx L(26)\approx 2.962\text{.}$

2. Take $f(x)=x^3-26\text{.}$ $x_1=3$ implies $x_2\approx 2.962\text{.}$

###### 30.

This question concerns finding zeros of the function

\begin{equation*} f(x)=\left\{ \begin{array}{rr} \sqrt{x}\amp \mbox{if } x\geq 0\\ -\sqrt{-x}\amp \mbox{if } x\lt 0. \end{array} \right. \end{equation*}
1. If the initial approximation to the zero, for $f(x)$ given above, is $x_1\text{,}$ what formula does Newton's method give for the next approximation?

2. The root of the equation $f(x)=0$ is $x=0\text{.}$ Explain why Newton's method fails to find the root no matter which initial approximation $x_1\not=0$ is used. Illustrate your explanation with a sketch.

1. Observe that $\ds f(x)=\mbox{sign} (x)\cdot \sqrt{|x|}$ and $\ds f^\prime(x)=\frac{1}{2\sqrt{|x|}}\text{,}$ so $x_{n+1}=-x_n\text{.}$

2. By induction, for all $n\in \mathbb{N}\text{,}$ $|x_{n}|=|x_1|\text{.}$

###### 31.
1. Suppose $k$ is a constant. Show that if we apply Newton's method to approximate the value of $\sqrt[5]{k}\text{,}$ we get the following iterative formula:

\begin{equation*} x_{n+1}=\frac{x_n}{5}\left( 4+\frac{k}{x_n^5}\right)\text{.} \end{equation*}
2. If $x_n=\sqrt[5]{k}\text{,}$ what is the value of $x_{n+1}\text{?}$

3. Take $x_1=2$ and use the formula in part (a) to find $x_2\text{,}$ an estimate of the value of $\sqrt[5]{20}$ that is correct to one decimal place.

1. Take $f(x)=x^5-k\text{.}$ Then $f^\prime (x)=5x^4$ and $\ds x_{n+1}=x_n-\frac{x_n^5-k}{5x_n^4}=\frac{4x_n^5+k}{5x_n^4}=\frac{x_n}{5}\left( 4+\frac{k}{x_n^5}\right)\text{.}$

2. $\ds x_{n+1}=\sqrt[5]{k}\text{.}$

3. $x_2=1.85\text{.}$ [MAPLE gives $\sqrt[5]{20}\approx 1.820564203\text{.}$]

###### 32.

Use Newton's method to find the second approximation $x_2$ of $\sqrt[5]{31}$ starting with the initial approximation $x_0=2\text{.}$

Solution

From $f(x)=x^5-31$ and $f'(x)=5x^4$ it follows that $\ds x_1=\frac{159}{80}$ and $\ds x_2=\frac{159}{80}-\frac{\left( \frac{159}{80}\right) ^5-31}{5\cdot \left( \frac{159}{80}\right) ^4}\approx 1.987340780\text{.}$ (Note: MAPLE gives $\sqrt[5]{31}=1.987340755\text{.}$)

###### 33.
1. Suppose $x_0$ is an initial estimate in Newton's method applied to the function $f(x)\text{.}$ Derive Newton's formula for $x_1\text{,}$ namely

\begin{equation*} x_1=x_0-\frac{f(x_0)}{f'(x_0)}\text{.} \end{equation*}

Support your derivation with a sketch showing a function $f(x)\text{,}$ with $x_0\text{,}$ $x_1$ and the line whose slope is $f'(x_0)$ clearly labeled.

2. Using one iteration of Newton's method with $\ds x_0=\frac{\pi }{2}$ approximate the $x$-coordinate of the point where the function $g(x)=\sin x$ crosses the line $y=x\text{.}$

Solution

The question is approximate the solution of the equation $F(x)=\sin x-x=0$ with $\ds x_0=\frac{\pi }{2}\text{.}$ Thus $\ds x_1=\frac{\pi }{2}-\frac{\sin \frac{\pi }{2}-\frac{\pi }{2}}{\cos \frac{\pi }{2}-1}=1\text{.}$ (Note: Clearly the solution of the given equation is $x=0\text{.}$ Newton's method with $x_0=\frac{\pi }{2}$ gives $x_7=0.08518323251\text{.}$)

###### 34.

The equation

\begin{equation*} 8x^3-12x^2-22x+25=0 \end{equation*}

has a solution near $x_1=1\text{.}$ Use Newton's Method to find a better approximation $x_2$ to this solution. Express your answer as a fraction.

$\ds x_2=1-\frac{-1}{-22}=\frac{21}{22}\text{.}$ Note: MAPLE gives $\ds \frac{21}{22}\approx 0.9545454545$ and approximates the solution of the equation as $0.9555894038$)
###### 35.

The tangent line to the graph $y=f(x)$ at the point $A(2,-1)$ is given by $y=-1+4(x-2)\text{.}$ It is also known that $f^{\prime\prime}(2)=3\text{.}$

1. Assume that Newton's Method is used to solve the equation $f(x)=0$ and $x_0=2$ is the initial guess. Find the next approximation, $x_1\text{,}$ to the solution.

2. Assume that Newton's Method is used to find a critical point for $f$ and that $x_0=2$ is the initial guess. Find the next approximation, $x_1\text{,}$ to the critical point.

1. $\ds x_1=2-\frac{-1}{4}=\frac{9}{4}\text{.}$

2. The question is to approximate a solution of the equation $f'(x)=0$ with the initial guess $x_0=2\text{,}$ $f'(2)=4\text{,}$ and $f"(2)=3$ given. Hence $x_1=2-\frac{4}{3}=\frac{2}{3}\text{.}$

###### 36.
1. Apply Newton's method to the equation $\ds \frac{1}{x}-a=0$ to derive the following algorithm for finding reciprocals:

\begin{equation*} x_{n+1}=2x_n-ax_n^2\text{.} \end{equation*}
2. Use the algorithm from part (a) to calculate $\ds \frac{1}{1.128}$ correct to three decimal places, starting with the first approximation $x_1=1\text{.}$

Solution
1. From $\ds f(x)=\frac{1}{x}-a$ and $\ds f'(x)=-\frac{1}{x^2}$ it follows that $\ds x_{n+1}=x_n-\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}=2x_n-ax_n^2\text{.}$

2. Note that $\ds \frac{1}{1.128}$ is the solution of the equation $\ds \frac{1}{x}-1.128=0\text{.}$ Thus $x_2=2-1.128=0.872\text{,}$ $x_3=2\cdot 0.872-1.128\cdot 0.872^2=0.886286848\text{,}$ and $x_4=0.8865247589\text{.}$ (Note: MAPLE gives $\ds \frac{1}{1.128}\approx 0.8865248227\text{.}$)

###### 37.
1. Apply Newton's method to the equation $\ds x^2-a=0$ to derive the following algorithm for the roots:

\begin{equation*} x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)\text{.} \end{equation*}
2. Approximate $\sqrt{2}$ by taking $x_1=2$ and calculating $x_2\text{.}$

$\ds x_2=\frac{3}{2}\text{.}$
###### 38.
1. State the formula for the linearization of $f$ at $a\text{,}$

2. Using linear approximation, approximate $\sqrt[4]{81.1}\text{.}$

3. Approximate $\sqrt[4]{81.1}$ using one iteration of Newton's method.

1. Take $f(x)=\sqrt[4]{x}$ with $a=81\text{.}$ $\sqrt[4]{81.1} \approx L(81.1)\approx 3.000925\text{.}$

2. Take $g(x)=x^4-81.1\text{.}$ $x_2= 3.000925\text{.}$

###### 39.

You seek the approximate value of $x$ which is near 1.8 for which $\ds \sin x=\frac{x}{2}\text{.}$ Your first guess is that $x\approx x_1=\frac{\pi }{2}\text{.}$ Use one iteration of Newton's method to find a better approximation to $x\text{.}$ Simplify your answer as far as possible.

Solution

$\ds x_2=\frac{\pi }{2}-\frac{1-\frac{\pi }{4}}{-\frac{1}{2}}=2\text{.}$ (Note: MAPLE estimates the positive solution of the equation $\ds \sin x=\frac{x}{2}$ as $1.895494267\text{.}$ Newton's method with the initial guess $\ds x_1=\frac{\pi }{2}$ gives $x_3\approx 1.900995594\text{.}$)

###### 40.
1. For the function $f(x)=x^3-3x+5$ use the Intermediate Value Theorem, and any other tools you need to determine intervals of length 1 each of which contains a root of $f\text{.}$

2. Pick one of the intervals found in part (a). Choose the left endpoint of this interval to be $x_0\text{.}$ Now, use this as a starting value to find two new iterations to the root of $f$ by using Newton's method. Determine from these whether Newton's method is working. Justify your answer carefully.

Solution
1. From $f'(x)=3(x^2-1)$ it follows that the critical numbers are $x=\pm 1\text{.}$ From $f(1)=3\text{,}$ $f(-1)=7\text{,}$ $\ds \lim _{x\to -\infty}f(x)=-\infty\text{,}$ and $\ds \lim _{x\to \infty}f(x)=\infty$ it follows that $f$ has only one root and that root belongs to the interval $(-\infty ,-1)\text{.}$ From $f(-2)=3>0$ and $f(-3)=-13\lt 0\text{,}$ by the Intermediate Value Theorem, we conclude that the root belongs to the interval $(-3,-2)\text{.}$

2. Let $x_0=-3\text{.}$ Then $\ds x_1=-3-\frac{-13}{504}=-\frac{1499}{504}\approx -2.974206349$ and $x_3\approx -2.447947724\text{.}$ It seems that Newton's method is working, the new iterations are inside the interval $(-3,-2)$ where we know that the root is. (Note: MAPLE estimates the solution of the equation $\ds x^3-3x+5=0$ as $x=-2.279018786\text{.}$)

###### 41.

Let $f(x)=x^3+3x+1\text{.}$

1. Show that $f$ has at least one root in the interval $\ds \left( -\frac{1}{2},0\right)\text{.}$ Explain your reasoning.

2. Use Newton's method to approximate the root that lies in the interval $\ds \left( -\frac{1}{2},0\right)\text{.}$ Stop when the next iteration agrees with the previous one to two decimal places.

Solution
1. The function $f$ is continuous on the closed interval $\ds \left[ -\frac{1}{2},0\right]$ and $\ds f\left( -\frac{1}{2}\right)=-\frac{5}{8}\lt 0$ and $f(0)=1>0\text{.}$ By the Intermediate Value Theorem, the function $f$ has at least one root in the interval $\ds \left( -\frac{1}{2},0\right)\text{.}$

2. Take $\ds x_1=-\frac{1}{3}\text{.}$ Then $\ds x_2=-\frac{1}{3}-\frac{-\frac{1}{27}-3\cdot \left( -\frac{1}{3}\right)+1}{3\left( \frac{1}{9}+1\right)} =-\frac{29}{90}\approx -.3222222222$ and $\ds x_3\approx -.3221853550\text{.}$ (Note: MAPLE estimates the solution of the equation $\ds x^3+3x+1=0$ as $x=-.3221853546\text{.}$)

###### 42.

In this question we investigate the solution of the equation $\ln x=-x^2+3$ on the interval $[1,3]\text{.}$

1. Explain why you know the equation has at least one solution on $[1,3]\text{.}$

2. Show that the equation has exactly one solution on $[1,3]\text{.}$

3. Use Newton's Method to approximate the solution of the equation by starting with $x_1=1$ and finding $x_2\text{.}$

1. Take $f(x)=\ln x+x^2-3\text{,}$ evaluate $f(1)$ and $f(3)\text{,}$ and then use the Intermediate Value Theorem.

2. Note that $\ds f^\prime (x)=\frac{1}{x}+2x>0$ for $x\in (1,3)\text{.}$

3. From $\ds f(1)=-2$ and $f^\prime (1)=3$ it follows that $\ds x_2=\frac{5}{3}\approx 1.66\text{.}$ [MAPLE gives 1.592142937 as the solution.]

###### 43.

In this question we investigate the positive solution of the equation $x^2+x=5-\ln x\text{.}$

1. Explain why you know the equation has at least one positive solution.

2. Show that the equation has exactly one positive solution.

3. Use Newton's Method to approximate the solution of the equation by starting with $x_1=1$ and finding $x_2\text{.}$

1. Use the Intermediate Value Theorem.

2. Use Rolle's Theorem.

3. $x_2=1.75\text{.}$

###### 44.

In this question we investigate the solution of the equation $2x=\cos x\text{.}$

1. Explain why you know the equation has at least one solution.

2. Show that the equation has exactly one solution.

3. Use Newton's Method to approximate the solution of the equation by starting with $x_1=0$ and finding $x_2\text{.}$

1. Take $f(x)=2x-\cos x\text{,}$ evaluate $\ds \lim _{x\to -\infty }f(x)$ and $\ds \lim _{x\to \infty }f(x)\text{,}$ and then use the Intermediate Value Theorem.

2. Note that $\ds f^\prime (x)=2+\sin x>0$ for $x\in \mathbb{R}\text{.}$

3. From $\ds f(0)=-1$ and $f^\prime (0)=2$ it follows that $\ds x_2=\frac{1}{2}\text{.}$ [MAPLE gives 0.4501836113 as the solution.]

###### 45.

In this question we investigate the solution of the equation $2x-1=\sin x\text{.}$

1. Explain why you know the equation has at least one solution.

2. Show that the equation has exactly one solution.

3. Use Newton's Method to approximate the solution of the equation by starting with $x_1=0$ and finding $x_2\text{.}$

1. Take $f(x)=2x-1-\sin x\text{,}$ evaluate $\ds \lim _{x\to -\infty }f(x)$ and $\ds \lim _{x\to \infty }f(x)\text{,}$ and then use the Intermediate Value Theorem.

2. Note that $\ds f^\prime (x)=2-\cos x>0$ for $x\in \mathbb{R}\text{.}$

3. From $\ds f(0)=-1$ and $f^\prime (0)=1$ it follows that $\ds x_2=1\text{.}$ [MAPLE gives 0.8878622116 as the solution.]

###### 46.

In this question we investigate the positive solution of the equation $e^x=2\cos x\text{.}$

1. Explain why you know the equation has at least one positive solution.

2. Show that the equation has exactly one positive solution.

3. Use Newton's Method to approximate the solution of the equation by starting with $x_1=0$ and finding $x_2\text{.}$

1. Take $f(x)=e^x-\cos 2x$ and then use the Intermediate Value Theorem.

2. Observe that $f^\prime(x)>0$ for all $x>0\text{.}$

3. $x_2=1\text{.}$

###### 47.

Consider the equation

\begin{equation*} x^6-x-1=0\text{.} \end{equation*}
1. Apply the Intermediate Value Theorem to the function $f(x)=x^6-x-1$ to prove that the given equation has a root greater than 1. Make sure that you justify why the function $f$ is continuous on its domain.

2. Use the derivative of the function $f(x)=x^6-x-1$ to prove that the given equation has only one root greater than 1. Call that root $a\text{.}$ Show all your work. Clearly explain your reasoning.

3. State Newton's Method.

4. Use Newton's Method with the initial approximation $x_1=1$ to find $x_2$ and $x_3\text{,}$ the second and the third approximations to the root $a$ of the equation $x^6-x-1=0\text{.}$ You may use your calculator to find those values. Show all your work. Clearly explain your reasoning.

5. WolframAlpha gives $a\approx 1.13472\text{.}$ Use your calculator to evaluate the number $|x_3 -1.13472|\text{.}$ Are you satisfied with your approximation. Why yes or why not?

1. Take $f(x)=x^6-x-1$ and then use the Intermediate Value Theorem.

2. $x_2=1.2$ and $x_3\approx 1.143575843\text{.}$

3. $|x_3-1.1347|\approx 0.008855843\text{.}$

###### 48.
1. State Rolle's theorem.

2. Use Rolle's theorem to prove that $f(x)$ has a critical point in $[0,1]$ where

\begin{equation*} f(x)=\sin\left(\frac{\pi x}{2}\right) -x^2\text{.} \end{equation*}
3. Set up the Newton's method iteration formula ($x_{n+1}$ in terms of $x_n$) to approximate the critical point You do not need to simplify.

$\ds x_{n+1}=x_n-\frac{\sin\left(\frac{\pi x_n}{2}\right)-x_n^2}{\frac{\pi}{2}\cos\left(\frac{\pi x_n}{2}\right)-2x_n}\text{.}$
###### 49.
1. State the Mean Value Theorem.

2. Using the Mean Value Theorem, prove that $f(x)$ has a critical point in $[0,1]$ where

\begin{equation*} f(x)=\cos\left(\frac{\pi x}{2}\right) +x\text{.} \end{equation*}
3. Set up the Newton's method iteration formula to approximate the critical point You do not need to simplify.

$\ds x_{n+1}=x_n-\frac{\cos\left(\frac{\pi x_n}{2}\right)+x_n}{\frac{-\pi}{2}\cos\left(\frac{\pi x_n}{2}\right)+1}\text{.}$
###### 50.
1. State the Intermediate Value Theorem.

2. State the Mean Value Theorem.

3. Use the Intermediate Value Theorem and the Mean Value Theorem to show that the equation $1+2x+x^3+4x^5=0$ has exactly one real root.

Hint
###### 51.

A function $h(x)$ is said to have a fixed point at $x=c$ if $h(c)=c\text{.}$ Suppose that the domain and range of a function $f(x)$ are both the interval $[0,1]$ and that $f$ is continuous on this domain, with $f(0)\not= 0$ and $f(1)\not=1\text{.}$

1. Prove that $f$ has at least one fixed point. That is, prove that $f(c)=c$ for some $c\in(0,1)\text{.}$

2. Suppose that $f^\prime(x)\lt 1$ for all $x\in(0,1)\text{.}$ Prove that $f$ has exactly one fixed point in $[0,1]\text{.}$

3. Use Newton's method to determine an iteration formula for the fixed point $x=c\text{.}$

1. Consider $g(x)=f(x)-x\text{.}$
2. $\ds x_{n+1}=x_n-\frac{f(x_n)-x_n}{f^\prime(x_n)-1}\text{.}$