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Calculus IChapter 4: Application of DerivativesSection 4-11

Linear Approximations — Question 8

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Question 8

Use a linear approximation to estimate 10.2\sqrt{10.2}.

  • (a) Define an appropriate function f(x)f(x) and choose a point aa close to 10.2 where f(a)f(a) is easy to compute.

  • (b) Find the linear approximation L(x)L(x) to f(x)f(x) at x=ax = a.

  • (c) Use the approximation to estimate 10.2\sqrt{10.2}, and compare it with the actual value.

Original worksheet page 1: question and worked solution for 4-11-008
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Question 8 - Solution

For f(x)=xf(x)=\sqrt{x}, use the nearby perfect square a=9a=9.

f(9)=3,f′(9)=16,L(x)=3+x−96.f(9)=3,\qquad f'(9)=\frac16,\qquad L(x)=3+\frac{x-9}{6}.

Therefore

10.2≈L(10.2)=3.2.\boxed{\sqrt{10.2}\approx L(10.2)=3.2.}

Since f″(x)=−1/(4x3/2)<0f''(x)=-1/(4x^{3/2})<0 for x>0x>0, the tangent line lies above the curve, so this is an overestimate.

For comparison, 10.2≈3.193743885\sqrt{10.2}\approx3.193743885; the error is approximately 0.0062561150.006256115.

Original worksheet page 2: question and worked solution for 4-11-008

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