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Video Example Find the linearization of the function f(x) = Vx + 1 at a = 3 and use it to approximate the numbers V3.97 and V4.02. Are these approximations overestimates or underestimates? Solution The derivative of f(x) = (x + 1)1/2 is f'(x) = 2(x +1 and so we have f(3) = 3 x and f'(3) = Putting these values into the equation L(x) = f(a) + f'(a)(x - a), we see that the linearization is L(x) = f(3) + f'(3)(x - 3) = | (x - 3) The corresponding linear approximation is Vx +1* (when x is near 3). In particular, we have V3.97 = (round to four decimal places) 4 and 4.02 (round to four decimal places).

Video Example Find the linearization of the function f(x) = Vx + 1 at a = 3 and use it to approximate the numbers V3.97 and V4.02. Are these approximations overestimates or underestimates? Solution The derivative of f(x) = (x + 1)1/2 is f'(x) = 2(x +1 and so we have f(3) = 3 x and f'(3) = Putting these values into the equation L(x) = f(a) + f'(a)(x - a), we see that the linearization is L(x) = f(3) + f'(3)(x - 3) = | (x - 3) The corresponding linear approximation is Vx +1* (when x is near 3). In particular, we have V3.97 = (round to four decimal places) 4 and 4.02 (round to four decimal places).

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter5: Exponential And Logarithmic Functions
Section5.5: Properties Of Logarithms
Problem 106E
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пріе Video Example ) Find the linearization of the function f(x) = Vx + 1 at a = 3 and use it to approximate the numbers 3.97 and V4.02. Are these approximations overestimates or underestimates? Solution The derivative of f(x) = (x + 1)1/2 is 1 f'(x) = 2(x +1) () and so we have f(3) = 3 X and f'(3) = Putting these values into the equation L(x) = f(a) + f'(a)(x - a), we see that the linearization is L(x) = f(3) + f '(3)(x – 3) = )(x - 3) The corresponding linear approximation is Vx + 1 2 (when x is near 3). In particular, we have 3.97 2 4 + (round to four decimal places) 4 and 4.02 z 4 (round to four decimal places). 4 The linear approximation is illustrated in the figure below. -1 3 We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for V3.97 and V4.02, but the linear approximation gives an approximation over an entire interval.
Transcribed Image Text:пріе Video Example ) Find the linearization of the function f(x) = Vx + 1 at a = 3 and use it to approximate the numbers 3.97 and V4.02. Are these approximations overestimates or underestimates? Solution The derivative of f(x) = (x + 1)1/2 is 1 f'(x) = 2(x +1) () and so we have f(3) = 3 X and f'(3) = Putting these values into the equation L(x) = f(a) + f'(a)(x - a), we see that the linearization is L(x) = f(3) + f '(3)(x – 3) = )(x - 3) The corresponding linear approximation is Vx + 1 2 (when x is near 3). In particular, we have 3.97 2 4 + (round to four decimal places) 4 and 4.02 z 4 (round to four decimal places). 4 The linear approximation is illustrated in the figure below. -1 3 We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for V3.97 and V4.02, but the linear approximation gives an approximation over an entire interval.
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