Skip to main content

Section 6.10 Linear Approximation and Newton's Method

These answers correspond to the problems in Section 3.5.

  1. \(\ds dV=\frac{72}{\pi^2}\text{.}\) Observe that \(\ds dr=\frac{1}{8\pi}\text{.}\)

  2. Observe that \(\Delta A=A(s+\Delta s)-A(s)=2s\Delta s+(\Delta s)^2\) and \(dA=2s\Delta s\text{.}\) See Figure 6.10.1.

    Figure 6.10.1. A square, an approximation, and the error

    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. (a) \(\ds \sqrt[3]{65}\approx L(65)=\frac{193}{48}\text{.}\) (b) Overestimate.

  5. 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. 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. 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. 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. (a) The linearization of the function \(g\) at \(a=2\) is \(L(x)=-4+3(x-2)\text{.}\) Thus \(g(2.05)\approx L(2.05)=-3.85\text{.}\) (b) 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. (a) \(\ds L(x)=1-\frac{x}{2}\text{.}\) (b) \(\ds \sqrt{0.9}\approx 1-\frac{9}{20}=\frac{11}{20}\text{.}\) (c) \(\ds y=-\frac{x}{2}+1\text{.}\) (d) See Figure 6.10.2.

    Figure 6.10.2. \(f(x)=\sqrt{1-x}\) and its tangent at \(x=0\)

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

  12. (a) \(L(x)=1+x\text{.}\) (b) \(\ds \sqrt{1.1}=f(0.05)\approx L(0.05)=1.05\text{.}\) (c) An over-estimate since \(f\) is concave-down. MAPLE gives \(\sqrt{1.1} \approx 1.048808848\text{.}\)

  13. (a) \(\ds L(x)=2+\frac{x}{12}\text{.}\) (b) \(\ds \sqrt[3]{7.95}\approx L(-0.05)=2-\frac{1}{240}=\frac{479}{240}\) 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. (a) For \(a\in \mathbb{R}\text{,}\) \(L(x) = (1+a)^{100} + 100(1+a)^{99}(x-a)\text{.}\) (b) \((1.0003)^{100} \approx L(0.0003) = 1.03\text{.}\) (c) Underestimate.

  15. (a) \(\ds y=\frac{x}{9}+3\text{.}\) (b) \(\ds \sqrt[3]{30}\approx \frac{1}{9}+3=\frac{28}{9}\text{.}\) (Note: MAPLE gives \(\ds \frac{28}{9}\approx 3.111111111\) and \(\sqrt[3]{30}\approx 3.107232506\text{.}\)) (c) See Figure 6.10.3.

    Figure 6.10.3. \(f(x)=\sqrt[3]{27+3x}\) and its tangent at \(x=0\)

  16. 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. Take \(f(x)=e^x\text{.}\) \(e^{-0.015} \approx 0.985\text{.}\) Underestimate.

  18. (a) \(L(x)=x-1\text{.}\) (b) 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. \(\ds L(x)=10+\frac{1}{300}(x-1000)\) 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. (a) 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 \(f(1.001)\approx L(1.001)=2+0.7\cdot 0.001=2.0007\text{.}\) (b) 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 \((0,\infty)\text{.}\) (c) 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. (a) \(\ds L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi }{6})\text{.}\) (b) By the Mean Value Theorem, for \(\ds x>\frac{\pi }{6}\) and some \(\ds c\in (\frac{\pi }{6},x)\text{,}\) \(\ds \frac{f(x)-f(\frac{\pi }{6})}{x-\frac{\pi }{6}}=\frac{\sin x-\frac{1}{2}}{x-\frac{\pi }{6}}=f^\prime (c)=\cos c\leq 1\text{.}\) Since \(\ds x-\frac{\pi }{6}>0\text{,}\) the inequality follows. (c) 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. (a) \(f(0.9) \approx L(0.9) = 5.2\text{,}\) \(f(1.1) \approx L(1.1) = 4.8\text{.}\) (b) Overestimates.

  23. (a) \(f(0.9) \approx L(0.9) = 2.8\text{,}\) \(f(1.1) \approx L(1.1) = 3.2\text{.}\) (b) Overestimates.

  24. (b) 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. (b) Take \(f(x)=\sqrt[3]{x}\text{,}\) \(x_0=1\text{,}\) \(x_1=-2\text{,}\) \(x_2=4\text{,}\) and \(x_3=-8\text{.}\) See Figure 6.10.4.

    Figure 6.10.4. Newton's Method fails: \(f(x)=\sqrt[3]{x}\) and \(x_0=1\)

    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. 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! See Figure 6.10.4.

  28. (a) \(\ds x_{n+1}=x_n-\frac{x_n^3-68}{3x_n^2}\text{.}\) (b) \(x_0=4\text{.}\) Underestimate.

  29. (a) \(\sqrt[3]{26}\approx L(26)\approx 2.962\text{.}\) (b) Take \(f(x)=x^3-26\text{.}\) \(x_0=3\) implies \(x_1\approx 2.962\text{.}\)

  30. (a) Observe that \(\ds f(x)=\mbox{sign} (x)\cdot \sqrt{|x|}\) and \(\ds f^\prime(x)=\frac{1}{2\sqrt{|x|}}\text{.}\) \(x_{n+1}=-x_n\text{.}\) (b) By induction, for all \(n\in \mathbb{N}\text{,}\) \(|x_{n}|=|x_1|\text{.}\)

  31. (a) Take \(f(x)=x^5-k\text{.}\) Then \(f^\prime (x)=5x^4\) and \(\ds x_{n+1}=x_n-\frac{x_n^5-k}{5x_n^4}=\frac{4x_n^5+k}{5x_n^4}=\frac{x_n}{5}\left( 4+\frac{k}{x_n^5}\right)\text{.}\) (b) \(\ds x_{n+1}=\sqrt[5]{k}\text{.}\) (c) \(x_2=1.85\text{.}\) [MAPLE gives \(\sqrt[5]{20}\approx 1.820564203\text{.}\)]

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

  35. (a) \(\ds x_1=2-\frac{-1}{4}=\frac{9}{4}\text{.}\) (b) 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. (a) From \(\ds f(x)=\frac{1}{x}-a\) and \(\ds f'(x)=-\frac{1}{x^2}\) it follows that \(\ds x_{n+1}=x_n-\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}=2x_n-ax_n^2\text{.}\) (b) 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. (b) \(\ds x_2=\frac{3}{2}\text{.}\)

  38. (b) Take \(f(x)=\sqrt[4]{x}\) with \(a=81\text{.}\) \(\sqrt[4]{81.1} \approx L(81.1)\approx 3.000925\text{.}\) (c) Take \(g(x)=x^4-81.1\text{.}\) \(x_2= 2.9997685\text{.}\)

  39. \(\ds x_2=\frac{\pi }{2}-\frac{1-\frac{\pi }{4}}{-\frac{1}{2}}=2\text{.}\) (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. (a) 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 \((-3,-2)\text{.}\) (b)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. (a) 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 \(\ds \left( -\frac{1}{2},0\right)\text{.}\) (b) 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. (a) Take \(f(x)=\ln x+x^2-3\text{,}\) evaluate \(f(1)\) and \(f(3)\text{,}\) and then use the Intermediate Value Theorem. (b) Note that \(\ds f^\prime (x)=\frac{1}{x}+2x>0\) for \(x\in (1,3)\text{.}\) (c) 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. (a) Use the Intermediate Value Theorem. (b) Use, for example, Rolle's Theorem. (c) \(x_2=1.75\text{.}\)

  44. (a) 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. (b) Note that \(\ds f^\prime (x)=2+\sin x>0\) for \(x\in \mathbb{R}\text{.}\) (c) 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. (a) 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. (b) Note that \(\ds f^\prime (x)=2-\cos x>0\) for \(x\in \mathbb{R}\text{.}\) (c) 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. (a) Take \(f(x)=e^x-\cos 2x\) and then use the Intermediate Value Theorem. (b) Observe that \(f^\prime(x)>0\) for all \(x>0\text{.}\) (c) \(x_2=1\text{.}\)

  47. (a) Take \(f(x)=x^6-x-1\) and then use the Intermediate Value Theorem. (d) \(x_2=1.2\) and \(x_3\approx 1.143575843\text{.}\) (e) \(|x_3-1.1347|\approx 0.008855843\text{.}\) {

  48. (c) \(\ds x_{n+1}=x_n-\frac{\sin\left(\frac{\pi x_n}{2}\right)-x_n^2}{\frac{\pi}{2}\cos\left(\frac{\pi x_n}{2}\right)-2x_n}\text{.}\)

  49. (c) \(\ds x_{n+1}=x_n-\frac{\cos\left(\frac{\pi x_n}{2}\right)+x_n}{\frac{-\pi}{2}\cos\left(\frac{\pi x_n}{2}\right)+1}\text{.}\) }

  50. (c) Proof by contradiction.

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