Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Determine the linear approximation \(L(x)\) at \(a\) of each function below. Then use \(L(x)\) to approximate the value of each function at the given \(x\)-value.

1. \(f(x) = \sqrt{x}, \ a=4, \ x=3\)

   Answer
   \(f(3) \approx L(3)=\frac{7}{4}\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} \sqrt{x} = \frac{1}{2\sqrt{x}} \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=4\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \sqrt{4} + \frac{1}{2\sqrt{4}} (x-4)\\ \amp = 2 + \frac{1}{4} (x-4) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(3) \approx L(3)= 2 + \frac{1}{4} (3-4) = \frac{7}{4} \end{equation*}

2. \(f(x) = \sqrt[3]{x}, \ a=8, \ x=9\)

   Answer
   \(f(9) \approx L(9)=\frac{25}{12}\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} x^{1/3} = \frac{1}{3x^{2/3}} \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=8\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = 8^{1/3} + \frac{1}{3(8^{2/3})} (x-8)\\ \amp = 2 + \frac{1}{12} (x-8) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(9) \approx L(9)= 2 + \frac{1}{12} (9-8) = \frac{25}{12} \end{equation*}

3. \(f(x) = \dfrac{1}{x}, \ a=5, \ x=5.3\)

   Answer
   \(f(5.3) \approx L(5.3)=\frac{47}{250}\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} \frac{1}{x} = -\frac{1}{x^2} \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=5\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \frac{1}{5} - \frac{1}{25} (x-5) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(5.3) \approx L(5.3)= \frac{1}{5} - \frac{1}{25} (0.3) = \frac{47}{250} \end{equation*}

4. \(f(x) = \dfrac{1}{x^{2}}, \ a=3, \ x=2.8\)

   Answer
   \(f(2.8) \approx L(2.8)=\frac{13}{135}\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} \frac{1}{x^2} = -\frac{2}{x^3} \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=3\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = \frac{1}{9} - \frac{2}{27} (x-3) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(2.8) \approx L(2.8)= \frac{1}{9} - \frac{2}{27} (0.2) = \frac{13}{135} \end{equation*}

5. \(f(x) = x^{2}+3, \ a=2, \ x=2.2\)

   Answer
   \(f(2.2) \approx L(2.2)=7.8\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} (x^2+3) = 2x \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=2\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = (4+3) + 4 (x-2)\\ \amp = 7 + 4 (x-2) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(2.2) \approx L(2.2)= 7+4(0.2) = 7+0.8 = 7.8 \end{equation*}

6. \(f(x) = (x-2)^{3}, \ a=3, \ x=3.1\)

   Answer
   \(f(3.1) \approx L(3.1)=1.3\)
   Solution

   We first differentiate:

   \begin{equation*} f'(x) = \diff{}{x} (x-2)^3 = 3(x-2)^2 \end{equation*}

   Next, use the formula for \(L(x)\) when \(a=3\text{:}\)

   \begin{equation*} \begin{split} L(x) \amp = f(a) + f'(a)(x-a)\\ \amp = 1 + 3 (x-3) \end{split} \end{equation*}

   Finally, use \(L(x)\) to approximate the given quantity:

   \begin{equation*} f(3.1) \approx L(3.1)= 1+3(0.1) = 1+0.3 =1.3 \end{equation*}

Find the linearization \(L(x)\) of \(f(x)=\ln (1+x)\) at \(a=0\text{.}\) Use this linearization to approximate \(f(0.1)\text{.}\)

Use linear approximation to estimate \((1.9)^3\text{.}\)

Show in detail that the linear approximation of \(\sin x\) at \(x=0\) is \(L(x)=x\) and the linear approximation of \(\cos x\) at \(x=0\) is \(L(x)=1\text{.}\)

Use \(f(x)=\sqrt[3]{x+1}\) to approximate \(\sqrt[3]{9}\) by choosing an appropriate point \(x=a\text{.}\) Are we over- or under-estimating the value of \(\sqrt[3]{9}\text{?}\) Explain.

We wish to estimate \(\sqrt[3]{9}\) using a linear approximation to \(f(x) = \sqrt[3]{x+1}\text{.}\)

The closest \(x\)-value for which we know the exact evaluation of \(f(x)\) is \(x=7\text{:}\)

\begin{equation*} f(7) = \sqrt[3]{8} = 2\text{.} \end{equation*}

We then choose \(a=7\) for our approximation, giving

\begin{equation*} \begin{split} L(x) \amp = f(7) + f'(7)(x-7)\\ \amp = f(7) + \frac{1}{3}\left(x+1\right)^{-2/3}\big\rvert_{x=7} \cdot (x-7) \\ \amp = 2 + \frac{1}{3}8^{-2/3} (x-7) \\ \amp = 2 + \frac{1}{12}(x-7) \end{split} \end{equation*}

So,

\begin{equation*} \sqrt[3]{9} = f(8) \approx L(8) = 2 + \frac{1}{12}(1) = \frac{25}{12}\text{.} \end{equation*}

This is a slight overestimation of the true value, since the slope of the tangent line to \(f\) is decreasing from \(x = 7\) to \(x = 8\text{.}\) To check our calculations, we compute

\begin{equation*} L(8) = 12 \approx 2.0833 \dots \end{equation*}

and the exact value

\begin{equation*} f(8) = 9 \approx 2.080008\dots \end{equation*}

The approximation is therefore accurate to 0.03. See the plots below of \(f(x) (x \geq -1)\) , and the corresponding linear approximation around \(x = 7\text{.}\)

Find the differential of the given function.

1. \(\ds f(x) = 2x^{2}\)

   Answer
   \(\ds dy = 4x dx\)
   Solution
   \(dy = f'(x)\,dx = 4x \,dx\)

2. \(\ds g(t) = t^{3}-t\)

   Answer
   \(\ds dy = (3t^{2}-1)dt\)
   Solution
   \(dy = g'(t)\,dt = (3t^2-1)\,dt\)

3. \(\ds f(t) = \sqrt{t+1}\)

   Answer
   \(\ds dy = \frac{1}{2\sqrt{t+1}} dt\)
   Solution
   \(dy = f'(t)\,dt = \dfrac{1}{2\sqrt{t+1}}\,dt\)

4. \(\ds p(q) = 2q^{3/2}+q^{1/2}\)

   Answer
   \(\ds dy = \frac{6q+1}{2\sqrt{q}}dq\)
   Solution
   \(dy = p'(q)\,dq = \dfrac{6q+1}{2\sqrt{q}}\,dq\)

5. \(\ds h(s) = s + \dfrac{2}{s}\)

   Answer
   \(\ds dy = \frac{s^{2}-2}{s^{2}} ds\)
   Solution
   \(dy = h'(s)\,ds= \dfrac{s^2-s}{s^2}\,ds\)

6. \(\ds p(q) = \frac{q-1}{q^{2}+1}\)

   Answer
   \(\ds dy = \frac{-q^{2}+2q+1}{(q^{2}+1)^{2}} dq\)
   Solution
   \(dy = p'(q)\,dq = \dfrac{-q^2+2q+1}{(q^2+1)^2}\,dq\)

7. \(\ds f(x) = \sqrt{3x^{2} - x}\)

   Answer
   \(\ds dy = \frac{6x-1}{2\sqrt{3x^{2}-x}}dx\)
   Solution
   \(dy = f'(x)\,dx = \dfrac{6x-1}{2\sqrt{3x^2-x}}\,dx\)

For the following functions \(f(x)\text{,}\) determine \(\Delta y\) and \(dy\) at the given values of \(a\) and \(\Delta x\text{.}\)

1. \(\ds f(x) = x^4\text{,}\) \(a=1\text{,}\) \(dx= \Delta x =1/2\)

   Answer
   \(\Delta y=-15/16\text{,}\) \(dy=2\)
   Solution

   The differential is given by

   \begin{equation*} dy = f'(x) dx \big\vert_{x=1} = 4(1)^3 \cdot \frac{1}{2} = 2\text{.} \end{equation*}

   The actual change in \(y\) is

   \begin{equation*} \Delta y = f(a+\Delta x) - f(a) = 1-\left(\frac{1}{2}\right)^4 = -\frac{15}{16}\text{.} \end{equation*}

2. \(\ds f(x) = \sqrt{x}\text{.}\) If \(a=1\) and \(\Delta x=1/10\)

   Answer
   \(\ds \Delta y=\sqrt{11/10}-1\text{,}\) \(dy=0.05\)

3. \(f(x) = \sin (2x)\text{.}\) If \(a=\pi\) and \(\Delta x=\pi/100\)

   Answer
   \(\ds \Delta y=\sin(\pi/50)\text{,}\) \(dy=\pi/50\)

For the functions (i) \(f(x)=x^{2}-1\) with \(x\) changing from 1 to 0.9 and (ii) \(f(x)=\frac{1}{x}\) with \(x\) changing from \(-1\) to \(-1.01\text{,}\) do the following:

1. Calculate the differential of \(f\text{.}\)

   Answer
   (i) \(\ds dy=2x dx\text{,}\) and (ii) \(\ds dy=-\frac{1}{x^{2}} dx\)
   Solution

   The differential of \(f(x) = x^2-1\) is given by

   \begin{equation*} df = f'(x)dx = 2xdx\text{.} \end{equation*}

   The differential of \(f(x)=1/x\) is given by

   \begin{equation*} df = f'(x)dx = \frac{-1}{x^2}dx\text{.} \end{equation*}
2. Use your results from part (a) to find the approximate change in \(y\) for the given change in \(x\text{.}\) Answer
   (i) \(\ds dy \approx -0.02\text{,}\) and (ii) \(\ds dy \approx -0.0098\)
   Solution

   Let \(f(x) = x^2-1.\) We approximate \(\Delta y\) between \(x=1\) and \(x=0.9\) by

   \begin{equation*} \begin{split} \Delta y \approx dy \amp = f'(x)dx \\ \amp =f'(1)(0.9-1) \\ \amp = 2(-0.1)= -0.2. \end{split} \end{equation*}

   Let \(f(x) = 1/x. \) We approximate \(\Delta y\) between \(x=-1\) and \(x=-1.01\) by

   \begin{equation*} \begin{split} \Delta y \approx dy \amp = f'(x)dx \\ \amp =f'(-1)(-1.01+1) \\ \amp = -1(-0.01)= 0.01. \end{split} \end{equation*}
3. Calculate the actual change in \(y\) for the given change in \(x\) and compare your results with that obtained in part (b). Answer
   (i) \(\ds \Delta y = -0.19\text{,}\) and (ii) \(\ds \Delta y = -0.0099\text{.}\)
   Solution

   Let \(f(x) = x^2-1.\) The exact change in \(y\) is given by

   \begin{equation*} \Delta y = f(0.9) - f(1) = (0.9^2-1)-(1-1) = -0.19\text{.} \end{equation*}

   Our estimate in part (b) was therefore accurate to 0.01.

   Let \(f(x) = 1/x. \) The exact change in \(y\) is given by

   \begin{equation*} \Delta y = f(-1.01) - f(-1) = (1/(-1.01))-(-1) = 0.00990..\text{.} \end{equation*}

   Our estimate in part (b) was therefore accurate to \(0.0001\text{.}\)

Use differentials to approximate the given quantity.

1. \(\ds \sqrt{49.5}\)

   Answer
   \(\ds \frac{197}{28}\)
   Solution

   We know that \(f(49) = \sqrt{49} = 7\text{,}\) and so let \(f(x) = \sqrt{x}\text{,}\) \(a = 49\) and \(dx = 0.5\text{.}\) Therefore, the differential \(df\) is

   \begin{equation*} df =\frac{1}{2\sqrt{49}} (0.5) = \frac{1}{28}\text{.} \end{equation*}

   Therefore, since \(df \approx \Delta f\text{,}\) we have that

   \begin{equation*} \Delta f = f(49.5) - f(49) = f(49.5) - 7 \approx \frac{1}{28}\text{.} \end{equation*}

   Rearranging, we find that

   \begin{equation*} f(49.5) \approx \frac{1}{28} + 7 = \frac{197}{28}\text{.} \end{equation*}

2. \(\ds \sqrt[3]{8.2}\)

   Answer
   \(\ds 2.042\)
   Solution

   Since we know that \(f(8) = \sqrt[3]{8} = 2\text{,}\) let \(f(x) = \sqrt[3]{x}\text{,}\) \(a=8\) and \(dx = 0.2\text{.}\) Then the differential \(df\) is

   \begin{equation*} df = \frac{1}{3x^{2/3}}\bigg\vert_{x=8} (0.2) = \frac{2}{30\cdot 4} =\frac{1}{60}\text{.} \end{equation*}

   Now if we let \(df \approx \Delta f\text{,}\) then

   \begin{equation*} \Delta f = f(8.2)- f(8) \approx \frac{1}{60} \implies f(8.2) \approx \frac{1}{60} + 2 = \frac{121}{60}\text{.} \end{equation*}

   Note: \(|\frac{121}{60} - \sqrt[3]{8.2}| \approx 0.0001\text{.}\)

3. \(\ds \sqrt{4.05} + \frac{1}{\sqrt{4.05}}\text{.}\)

   Answer
   \(\ds 2.509\)
   Solution

   Let \(f(x) = \sqrt{x} + \dfrac{1}{\sqrt{x}}\text{.}\) The differential, \(df\text{,}\) of \(f\) is given by

   \begin{equation*} df=f'(x)\,dx = \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right)dx\text{.} \end{equation*}

   So we can evaluate exactly both \(f(4)\) and \(f'(4)\text{.}\) We can now estimate \(f(4.05) = \sqrt{4.05} + \dfrac{1}{\sqrt{4.05}}\text{:}\)

   \begin{equation*} \begin{split} \Delta f = f(4.05) - f(4) \amp \approx f'(4) (0.05) f(4.05) \\\amp \approx f'(4) (0.05) + f(4) \\ \amp \approx \left(\frac{1}{4} - \frac{1}{16}\right) (0.05) + \left(2+ \frac{1}{2}\right) \\ \amp \approx \frac{3}{16}(0.05)+\frac{5}{2} \approx 2.509375. \end{split} \end{equation*}

   We can compare this value to the exact value

   \begin{equation*} f(4.02)=2.509365 \dots \end{equation*}

   We see that this approximation is accurate to around \(10^{-6}\text{!}\)

Approximate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter \(40\) meters. You may use the fact that the volume of a sphere of radius \(r\) is \(V =(4/3)\pi r^3\text{,}\) where in this example, \(dr=0.02\text{.}\)

It is determined that a certain country's gross domestic product (GDP) can be approximated by

where \(f(x)\) is measured in millions of dollars and \(x\) is the capital expenditure in billions of dollars. Approximate the change in GDP if the country's capital expenditure changes from $200 million to $210 million.

A major supermarket determines that their yearly profit \(P(q)\) is related to the amount \(q\) spent on advertising by

where both \(P(q)\) and \(q\) are measured in thousands of dollars. Approximate the change in profits when advertising expenditure is increased from $30,000 to $32,000.

A bank determines that the number \(N(t)\) of loans issued over the course of one year is related to the interest rate \(r\) by

where \(N\) is measured in millions. Approximate the change in the number of loans the bank issues when the interest rate is increased from 10% to 10.5%.

The supply equation for a certain product is given by

where \(q\) is the quantity supplied and \(p\) is the unit price in dollars. Approximate the change in price when the quantity supplied is increased from 10,000 to 10,200 units.

The edges of a cube are measured to be 12 cm in length, with a maximum possible error of 0.02 cm. What is the maximum possible error that could occur when calculating the volume of the cube?

A wooden box with a lid is lacquered to an even thickness of 0.04 cm. If the edges of the box measure 0.5 m, then calculate the approximate amount of lacquer required.

A dome of radius 20 m is to be coated with a layer of paint. What is the approximate amount of paint needed if the coat is to be 0.05 cm thick? Note that the volume of a dome of radius \(r\) is \(V = \frac{2}{3}\pi r^{3}\text{.}\)

True or false: If \(A=f(x)\text{,}\) then the percentage change in \(A\) is

Explain your answer.

The quarterly profit of a certain manufacturer is given by

million dollars, where \(q\) is measured in tens of thousands of units. The expected number of units sold over the next quarter is 320,000, with a maximum error of 18%. Determine the maximum error in the expected profit.

The demand equation for a certain product is given by

where \(p\) is the unit price in dollars and \(q\) is the quantity demanded each year, measured in thousands of units. It is expected that the demand will be 2000 units for the year, with a maximum error of 10%. What is the maximum error in the predicted price?

Suppose a monthly mortgage payment \(P\text{,}\) in dollars, is computed using the formula

where \(r\) is the interest rate per year.

1. Find the differential of \(P\text{.}\)

   Answer
   \(\ds dP = \dfrac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}} dr\)
   Solution

   We first differentiate:

   \begin{equation*} P'(r) = \diff{}{r}\left(\frac{10,000 r}{1-\left(1+\frac{r}{12}\right)^{-360}}\right) = \dfrac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}} \end{equation*}

   Hence,

   \begin{equation*} dP = \diff{}{r}\left(\frac{10,000 r}{1-\left(1+\frac{r}{12}\right)^{-360}}\right) = \frac{10,000\left(1-(1+\frac{r}{12})^{-360}-30r(1+\frac{r}{12})^{-361}\right)}{\left(1-\left(1+\frac{r}{12}\right)^{-360}\right)^{2}}\,dr \end{equation*}

2. Approximately how much more will the monthly mortgage payments be if the interest rate increases from the present rate of 3% per year to 3.2 % per year? From 3% to 3.3% per year? To 3.4% per year? To 3.5% per year?

   Answer
   $12.90,$19.40,$25.89,$32.36
   Solution

   Let \(r=0.03\) and \(dr = 0.002\text{.}\) Then \(dP \approx 12.9\text{.}\) When \(dr = 0.003\text{,}\) \(dP \approx 19.4\text{,}\) and so on. We find that, when the interest rate increases to 3.2% per year, 3.3% per year, 3.4% per year and 3.5% per year, the monthly mortgage payments increase by about $12.90, $19.40, $25.89 and $32.36, respectively.

Suppose $10,000 is deposited into an account that pays interest at the rate \(r\)/year compounded monthly. Then the account balance at the end of 10 years is given by

1. Find the differential of \(A\text{.}\)

   Answer
   \(dA = 100,000\left(1+\frac{r}{12}\right)^{119}dr\)
   Solution

   We first differentiate:

   \begin{equation*} A'(r) = 100,000\left(1+\frac{r}{12}\right)^{119} \end{equation*}

   Therefore,

   \begin{equation*} dA = 100,000\left(1+\frac{r}{12}\right)^{119}\,dr\text{.} \end{equation*}

2. Approximately how much more would the account be worth at the end of 10 years with an interest rate of 1.1% per year instead of 1%? 1.2% per year instead of 1%? 1.3% per year instead of 1%?

   Answer
   $110,$220,$331
   Solution

   When \(r = 0.01\) and let \(dr = 0.001\text{,}\) we compute \(dA = 110.42\text{.}\) Hence, the account will be worth approximately $110.42 more after 10 years when the interest rate is raised to 1.1%. Similarly, we find that when the interest rate is raised to 1.2$ and 1.3%, the account is worth approximately $220 and $331 more, respectively.

Suppose $2000 per month is deposited into an account that pays interest at the rate \(r\)/year compounded monthly. Then the account balance at the end of 25 years is given by

dollars.

1. Find the differential of \(S\text{.}\)

   Answer
   \(dS =24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right)dr\)
   Solution

   We first differentiate:

   \begin{equation*} S'(r) = 24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right) \end{equation*}

   Therefore,

   \begin{equation*} dS = 24,000\left(\dfrac{300r(1+\frac{r}{12})^{299}(\frac{1}{12})-(1+\frac{r}{12})^{300}+1}{r^{2}} \right)\,dr\text{.} \end{equation*}

2. Approximately how much more would the account be worth at the end of 25 years with an interest rate of 1.6%/year instead of 1.5%? 1.7%/year instead of 1.5%? 1.8%/year instead of 1.5%?

   Answer
   $9617; $19,235; $28,852
   Solution

   When \(r=0.015\) and \(dr = 0.001\text{,}\) we compute \(dS = 9617\text{.}\) That is, the account will be worth approximately $9617 more after 25 years if the interest rate is raised to 1.6% per year. Similarly, the account will be worth about $19,235 and $28,852 more if the interest rate is raised to 1.7% and 1.8% per year, respectively.

Apply Newton's Method using the following steps: First, determine the function given the radical. Second, approximate the given radical with three iterations based on the given initial guess.

1. \(\ds \sqrt{3}\text{,}\) \(x_{0}=1.5\)

   Answer
   \(f(x)=x^{2}-3\text{,}\) 1.732051
   Solution

   We need to find a function \(f\) such that \(f(x) = 0 \iff x = \sqrt{3}\text{.}\) We cannot take \(f(x) = x-\sqrt{3}\) (since we are trying to approximate \(\sqrt{3}\text{!}\)), let \(f(x) = x^2-3\) and search for the positive root. Apply Newton's Method using:

   \begin{equation*} x_{k+1} = x_k - \frac{x_k^2-3}{2x_k}, k = 0,1,2,3,..\text{.} \end{equation*}

   The first three iterations are shown in the table below.

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.5 - \dfrac{1.5^2-3}{2(1.5)} =1.75 \\ 2 \amp 1.75 - \dfrac{1.75^2-3}{2(1.75)}=1.73214 \\ 3 \amp 1.73214 - \dfrac{1.73214^2-3}{2(1.73214)} = 1.73205 \end{array} \end{equation*}

   Hence, \(\sqrt{3} \approx 1.73205\text{.}\) Note: \(|1.72305-\sqrt{3}| \approx 0.009\text{.}\)

2. \(\ds \sqrt{7}\text{,}\) \(x_{0}=2.5\)

   Answer
   \(f(x) = x^{2}-7\text{,}\) 2.645751
   Solution

   Suppose \(f(x) = x^2-7\text{.}\) Then \(f(x) = 0\) if and only if \(x = \pm \sqrt{7}\text{.}\) So to approximate \(\sqrt{7}\text{,}\) we can use Newton's Method with \(f(x) = x^2-7\) and search for the positive root. The iterative formula is thus:

   \begin{equation*} x_{n+1} = x_n - \frac{x_n^2-7}{2x_n}, n = 0,1,2,3,..\text{.} \end{equation*}

   The first three iterations are shown in the table below, taking \(x_0 = 2.5\text{.}\)

   \begin{equation*} \begin{array}{ll} n \amp x_n \\ \hline 0 \amp 2.5 \\ 1 \amp 2.5 - \dfrac{2.5^2-7}{2(2.5)} = 2.65 \\ 2 \amp 2.65 - \dfrac{2.65^2-7}{2(2.65)}= 2.64575471... \\ 3 \amp 2.645751... \end{array} \end{equation*}

   Therefore, \(\sqrt{7} \approx 2.64575\text{.}\)

3. \(\ds \sqrt[3]{14}\text{,}\) \(x_{0}=2.5\)

   Answer
   \(f(x)= x^{3}-14\text{,}\) 2.410142
   Solution

   Suppose \(f(x) = x^3-14\text{.}\) Then \(f(x) = 0\) if and only if \(x = \sqrt[3]{14}\text{.}\) Therefore, we use Newton's Method with the iterative formula:

   \begin{equation*} x_{k+1} = x_{k} - \frac{x_k^3-14}{3x_k^2}, k = 0,1,2,3,..\text{.} \end{equation*}

   Taking an initial guess \(x_0=2.5\text{,}\) the first three iterations are shown in the table below.

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2.5 \\ 1 \amp 2.5 - \dfrac{2.5^3-14}{3(2.5)^2} =2.4133 \\ 2 \amp 2.41015\\ 3 \amp 2.41014 \end{array} \end{equation*}

   Hence, \(\sqrt[3]{14} \approx 2.4101\text{.}\)

Apply Newton's Method to approximate the root of each function within the given interval, accurate to three decimal places.

1. \(f(x)=e^{-x}-x\text{,}\) \((0,1)\text{.}\)

   Answer
   0.567143
   Solution

   We wish to approximate a root of \(f(x) = e^{-x}-x\) in the interval \((0,1)\text{.}\) Notice that \(f(0) = 1 > 0\) and \(f(1)=-0.632... \lt 0\) and so by the Intermediate Value Theorem, \(f\) must contain at least one root in this interval. Our Newton iterations will be of the form

   \begin{equation*} x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = x_k + \frac{e^{-x}-x}{e^{-x}+1}, k = 0,1,2,3,..\text{.} \end{equation*}

   For our initial guess, take \(x_0 = 0.5\text{.}\)

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 0.5 \\ 1 \amp 0.566311\\ 2 \amp 0.567143\\ 3 \amp 0.567143 \end{array} \end{equation*}

   We stop after 3 iterations since the method has converged on a root. Thus, \(x = 0.567143\) is our approximation to a root of \(f\text{.}\)

2. \(f(x)=e^{x} - \frac{1}{x}\text{,}\) \((0,1)\text{.}\)

   Answer
   0.567143
   Solution

   Notice that \(x=1\) is a root of \(f(x)\text{.}\) We can check if there is a root in the interval \((0,1)\) using Newton's Method with \(x_0=0.5\text{:}\)

   \begin{equation*} x_{k+1} = x_k - \frac{e^{x_k}-\frac{1}{x_k}}{e^{x_k} +\frac{1}{x^2_k}}, k = 0,1,2,3,..\text{.} \end{equation*}

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 0.5 \\ 1 \amp 0.562187\\ 2 \amp 0.56712\\ 3 \amp 0.567143 \end{array} \end{equation*}

   Hence, \(x=0.567\) is a root of \(f\) to three decimal places.

3. \(f(x)=\ln (x^{2})\text{,}\) \((6,7)\text{.}\)

   Answer
   None
   Solution

   The function \(f(x) = \ln(x^2)\) has roots

   \begin{equation*} x^2= 1 \implies x = \pm 1 \end{equation*}

   and so has no roots in the interval \((6,7)\text{.}\)

Sketch the graphs of \(f\) and \(g\) on the same Cartesian coordinate system. Based on the sketch, choose an initial value to approximate the \(x\)-coordinate of the intersection point(s) of the two graphs. Then apply Newton's Method to determine the \(x\)-coordinate of the intersection point accurate to within two decimal places.

1. \(\ds f(x) = \sqrt{x}\text{,}\) \(g(x)=-0.5x+2\)

   Answer
   1.528
   Solution

   From the sketch below, we see that \(x_0 = 1.5\) is a reasonable starting guess for the intersection point of \(f\) and \(g\text{:}\)

   

   Let \(h(x) = f(x) - g(x)\text{.}\) Then Newton's Method applied to the function \(h(x)\) will approximate

   \begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

   Newton's formula is thus

   \begin{equation*} x_{k+1} = x_k - \frac{\sqrt{x_k} +0.5x_k - 2}{\frac{1}{2x_k}-0.5} k = 0,1,2,..\text{.} \end{equation*}

   The first three iterations are shown below:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.52781\\ 2 \amp 1.52786\\ 3 \amp 1.52786 \end{array} \end{equation*}

   Hence, the intersection point of \(f\) and \(g\) is approximately \(1.528\text{.}\)

2. \(\ds f(x)=e^{-x^{2}}\text{,}\) \(g(x)=x^{2}\)

   Answer
   0.753
   Solution

   From the sketch below, we see that \(x_0 = 1\) is a reasonable starting guess for the intersection point of \(f\) and \(g\text{:}\)

   

   Let \(h(x) = f(x) - g(x)\text{.}\) Then Newton's Method applied to the function \(h(x)\) will approximate

   \begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

   Newton's formula is thus

   \begin{equation*} x_{k+1} = x_k - \frac{e^{-x_k^2} - x^2}{-2x_ke^{-x_k^2} - 2x_k} k = 0,1,2,..\text{.} \end{equation*}

   The first three iterations are shown below:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1 \\ 1 \amp 0.768941\\ 2 \amp 0.753185\\ 3 \amp 0.753089 \end{array} \end{equation*}

   Hence, the intersection point of \(f\) and \(g\) is approximately \(0.753\text{.}\)

3. \(\ds f(x)=\ln x\text{,}\) \(g(x)=2-x\)

   Answer
   1.557
   Solution

   From the sketch below, we see that \(x_0 = 2\) is a reasonable starting guess for the intersection point of \(f\) and \(g\text{:}\)

   

   Let \(h(x) = f(x) - g(x)\text{.}\) Then Newton's Method applied to the function \(h(x)\) will approximate

   \begin{equation*} h(x) = 0 \implies f(x) - g(x) = 0 \implies f(x) = g(x)\text{.} \end{equation*}

   Newton's formula is thus

   \begin{equation*} x_{k+1} = x_k - \frac{ln(x_k) -2 + x_k}{\frac{1}{x_k} + 1} k = 0,1,2,..\text{.} \end{equation*}

   The first three iterations are shown below:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2 \\ 1 \amp 1.5379\\ 2 \amp 1.5571\\ 3 \amp 1.55715 \end{array} \end{equation*}

   Hence, the intersection point of \(f\) and \(g\) is approximately \(1.557\text{.}\)

For each of the given functions \(f(x)\text{,}\) show that \(f(x)=0\) has a root between the given \(x\)-values. Use Newton's Method to find the zero(s). Hint

1. \(f(x)=3x^{2}-9x-11\) between \(x=-1\) and \(x=0\text{;}\) and between \(x=3\) and \(x=4\text{.}\)

   Solution

   The function \(f(x)\) is a polynomial (and hence continuous everywhere), and so we can apply IVT. We first investigate the root between \(x=-1\) and \(x=0\text{.}\) Since \(f(-1) = 3 +9 - 11 > 0\text{,}\) and \(f(0) = -11 \lt 0\text{,}\) by the Intermediate Value Theorem, \(f\) must contain at least one root in the interval \((-1,0)\text{.}\) Let \(x_0 = -0.5\) and

   \begin{equation*} x_{k+1} = x_k - \frac{3x_k^2-9x_k-11}{6x_k+9}, k = 0,1,2,3,..\text{.} \end{equation*}

   Then:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp -0.5 \\ 1 \amp -0.979167 \\ 2 \amp -0.932861 \\ 3 \amp -0.93242 \\ 4 \amp -0.93242 \end{array} \end{equation*}

   We conclude that \(x = -0.93242\) approximates a root of \(f(x)\text{.}\) We see that \(f(x)\) must also have a root on the interval \((3,4)\text{,}\) since

   \begin{equation*} f(3)= 3(9) - 9(3) -11 \lt 0, \ \text{ and } \ f(4) = 3(16)-9(4)-11 > 0\text{.} \end{equation*}

   So, take the same iterative formula as above, but now take \(x_0 = 3.5\text{:}\)

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 3.5 \\ 1 \amp 3.97917 \\ 2 \amp 3.93286 \\ 3 \amp 3.93242\\ 4 \amp 3.93242 \end{array} \end{equation*}

   Hence, we conclude that \(x = 3.93242\) also approximates a root of \(f(x)\text{.}\)

2. \(f(x)=x^{3}-x-1\) between \(x=1\) and \(x=2\text{.}\)

   Solution

   The function \(f(x)\) is a polynomial (and hence continuous everywhere), and so we can apply IVT. Since \(f(1) = 1-1-1 = -1 \lt 0\) and \(f(2) = 8-2-1=5 > 0\text{,}\) by the Intermediate Value Theorem, \(f\) must contain at least one root in the interval \((1,2)\text{.}\) To find this root, take \(x_0=1.5\) and the Newton iterations

   \begin{equation*} x_{k+1} = x_k - \frac{x_k^3-x_k-1}{3x_k^2-1}, k = 0,1,2,3,..\text{.} \end{equation*}

   Then:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.34783 \\ 2 \amp 1.3252 \\ 3 \amp 1.32472 \\ 4 \amp 1.32472 \end{array} \end{equation*}

   We stop after 4 iterations and conclude that \(x = 1.32472\) approximates a root of \(f\text{.}\)

3. \(f(x)=x^{4}-4x^{3} + 10\) between \(x=1\) and \(x=2\text{.}\)

   Solution

   The function \(f(x)\) is a polynomial (and hence continuous everywhere), and so we can apply IVT. We compute

   \begin{equation*} f(1) = 1-4+10 > 0, \ \text{ and } \ f(2) = 16 - 4(8) + 10 \lt 0\text{.} \end{equation*}

   Therefore, by the Intermediate Value Theorem, \(f\) must have at least one root on the interval \((1,2)\text{.}\) Let \(x_0 = 1.5\) and

   \begin{equation*} x_{k+1} = x_k - \frac{x_k^4-4x_k^3+10}{4x_k^3-12x_k^2}, k = 0,1,2,3,..\text{.} \end{equation*}

   Then:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 1.5 \\ 1 \amp 1.61574 \\ 2 \amp 1.6118 \\ 3 \amp 1.61179 \\ 4 \amp 1.61179 \end{array} \end{equation*}

   We conclude that \(x=1.61179\) approximates a root of \(f\text{.}\)

Consider \(f(x)=x^3-x^2+x-1\text{.}\)

1. Using initial approximation \(x_0=2\text{,}\) find \(x_4\text{.}\)

   Answer
   \(x_4\approx 1.00022\ldots\)
   Solution

   We are given \(f(x) = x^3-x^2+x-1\text{.}\) Then the iterative formula required for Newton's Method is

   \begin{equation*} x_{k+1} = x_k - \frac{x_k^3-x_k^2+x_k-1}{3x_k^2-2x_k+1}, k = 0,1,2,3,..\text{.} \end{equation*}

   Taking \(x_0 = 2\text{,}\) we find:

   \begin{equation*} \begin{array}{ll} k \amp x_k \\ \hline 0 \amp 2 \\ 1 \amp 1.44444 \\ 2 \amp 1.13057 \\ 3 \amp 1.01498 \\ 4 \amp 1.00022 \end{array} \end{equation*}

   Therefore, \(x_4 = 1.00022\text{.}\)

2. What is the exact value of the root of \(f\text{?}\) How does this compare to our approximation \(x_4\) in part (a)?

   Answer
   \(x=1\) is the root of \(f\text{.}\) Our approximation in part (a) was correct to 3 decimal places.
   Solution

   By inspection, we see that \(x=1\) is the exact value of a root of \(f\text{.}\) So the error in our approximation is

   \begin{equation*} |x_4 - 1| = |1.00022 - 1| = 0.00022\text{.} \end{equation*}

   So our approximation \(x_4\) is correct to 3 decimal places.

3. What would happen if we chose \(x_0=0\) as our initial approximation?

   Answer
   \(x_1=1\text{.}\) The root is found in one iteration of Newton's Method.
   Solution

   If we take \(x_0 = 0\text{,}\) we find that

   \begin{equation*} x_1 = 0 - \frac{-1}{1} = 1\text{,} \end{equation*}

   and so the method converges to the exact root after only one iteration.

Consider \(f(x)=\sin x\text{.}\) What happens when we choose \(x_0=\pi/2\text{?}\) Explain.

Let \(f(x) = \sin x\text{.}\) Then we know that the roots of \(f(x)\) are \(n\pi\text{,}\) for all integers \(n\text{.}\)

If we tried to apply Newton's Method to \(f\) taking \(x_0=\frac{\pi}{2}\text{,}\) we can see graphically that this will not converge to a root:

We can also argue algrabraically: the iterative formula required is

\begin{equation*} x_{k+1} = x_{k} - \frac{\sin x_k}{\cos x_k}, k = 0,1,2,3,..\text{.} \end{equation*}

and so if \(x_0 = \frac{\pi}{2}\text{,}\) then \(\cos(x_0) = 0\text{.}\) Hence, \(x_1\) would be undefined.

Suppose a company purchases $6000 worth of new equipment which will be used for the next 3 years. The investment is expected to yield returns of $2,000 at the end of the first year, $3500 at the end of the second year and $1000 at the end of the third year. What is the internal rate of return on this investment?

Executives of a certain company are contemplating the purchase of $100,000 worth of equipment, which would be in use over the next 4 years. The investment is expected to yield returns of $20,000 at the end of the first year, $30,000 at the end of the second year, $45,000 at the end of the third year and $15,000 at the end of the fourth year. What is the internal rate of return on this investment?

A first time home buyer borrows $250,000 from a bank to finance the purchase of a house. Interest is computed at the end of each month at a rate of 12\(r\) per year on the unpaid balance. If the home buyer repays the loan in equal monthly instalments of $2226.10 over the next 10 years, what is the rate of interest charged by the bank?

A down payment of 10% is made towards a purchase of $10,000. Financing for this purchase is available with monthly payments of $255.50 over 4 years. What is the rate of interest charged?

Suppose the demand equation for a certain product is given by

where \(p\) is the unit price in dollars and \(q\) is the quantity demanded in units of hundreds. The corresponding supply equation is given by

dollars. What is the equilibrium point?

A certain company determines that the demand equation for a product is

where \(p\) is the unit price in dollars and \(q\) is the quantity demanded in units of a thousand. If the weekly total cost function associated with the production of this product is

what is the break-even level(s) of operation for the company?