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Approximate 11 k k 1 f 0 solution where k

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Unformatted text preview: gives (x) of the function b. FindThe second which valuesxofof )does (this 1)(1 x)linearizationnear xunderestimates of the [3 points] For derivative L( f (k the function f ( + x)k(12+ so)k give = 0, where k is a positive constant. 12. 0. the local linearization is ) x = k k − local = − , x k √ f (x actual you of f to use L(x) your work.) ) = (1 + x) of the number 1.1. What number should k be, and Supposevaluewnat (x)? (Show to find an√ approximation ! what number should x be? Approximate (1.1.= k (k − 1). f 0) Solution: where k linearization gives underestimates of the actual value when f (0) > ! near x = 0, The localis a positive constant. 0. The second derivative is f (x) = k (k − 1)(1 1 x)k−2 , so + Since k > 0,The derivative iswhen) theksecond kfactor is positive, which is when k > 1. = 0 Solution: this is positive f (x = (1 + x) − , so the slope of the tangent line at x is f (0) = k (k − 1). f (0) = k. √ c. [2Since k >Suppose you want when the(second factor approximation of the number 1.1.1. points] 0, this is positive to use L x) to find an is positive, which is when k > k = 1, the tangent line passes through the point (0, 1). Therefore, the Since f (0) = 1 What number should k be, and what number should x be? point-slope formula shows that the equation of the tangent line is √ √ Solution: If k = 1 and x = 0.1, then f (0.1) = 1.1, so L(1.1) gives an approximation √ c. [2 points] Suppose2you want to use L(x)= kx + 1. approximation of the number 1.1. y to find an of .1. What1number should k be, and what number should x be? ! √ Solution: If k = 1 and x = 0.1, then f (0.1) = 1.1, so L(1.1) gives an approximation 2 √ √ d. [2of 1.1. Approximate 1.1 using L(x). points] b. [3 points] For which values of k does this local linearization give underestimates of the ! √ actual value of f (x)? (Show your work.) 1.1 ≈ L(0.1) = 1.05. Solution: If k and x are as above, then √ ! 1 d. [2Solution: Approximate 1.1 usinggives).underestimates of the actual value when f (0) > points] The local linearization L(x √ 0. linear approximation to estimate the k − number 13. Use aThe second derivative is f (x) = k (given1)(1 + x)k−2 , so e. [2Solution: What and x are asin the approximationL(0.1)part.05. points] If k is the error above, then 1.1 ≈ from = 1 (d)? (a) (8.06)2/3 f (0) = k (k − 1). Solution: e. [2Since k > 0, this is positive in the the second factor is positive, which is when k > 1. points] What is the error when approximation from part (d)? √√ 1− c. [2 points] Suppose you wantactual value) minus the approximate valuethe number 1.1 .1. Solution: The error is the to use L(x to find an approximation of which is What number should k be, and what number should x be? 1.05 ≈ −0.00119. √ √ 1 √ Solution: Solution: If k error is x actual value minus 1 1, so L(1.1) value which is 1.1 − (b) √99.8 The = 2 andthe= 0.1, then f (0.1) = the .approximate gives an approximation of Solution: 1− 1.05 ≈.1. 0.00119. ! 1For...
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This note was uploaded on 11/19/2012 for the course ECON CALC 1 taught by Professor Fisherselin during the Fall '11 term at NYU.

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