University:
Kentucky State University
Course:
MAT 115 | College Algebra
Academic year:

2024
Views:

457

Pages:

5
Author:

Sean Horton

Text from the Image: §3.10 Linear Approximations. curve y=f(x) with a tangent line at point (a, f(a)) The tangent line at (a, f(a)) is used as an approximation to the curve y=f(x) when x is near a. .: f(x) ≈ L(x) = f'(a)(x-a) + f(a) for all x near a. "Linearization" Example: Find the linearization of the function f(x) = √(x+3) at a = 1, and use it to approximate the numbers √3.98 and √4.05. Soln: f(1) = √(1+3) = 2, f'(x) = 1/(2√(x+3)) ⇒ f'(1) = 1/4 ∴ Linearization: L(x) = f'(1)(x-1) + f(1) = 1/4(x-1) + 2 = 1/4x + 7/4 Now, f(x) ≈ L(x) ⇒ √(x+3) ≈ 1/4x + 7/4 when x is near 1. (i) √3.98 = √(0.98+3) ≈ 0.98/4 + 7/4 = 1.995 (ii) √4.05 = √(1.05+3) ≈ 1.05/4 + 7/4 = 2.0125 Example: Verify the given Linear approximation at a = 0. ① sinx ≈ x. (Exercise) ② cosx ≈ 1. (Exercise) ③ tanx ≈ x. (Exercise) ④ e^(x)cosx ≈ 1+x. Soln: Let f(x) = e^(x)cosx, a = 0, f(0) = e^(0)cos0 = 1, f'(x) = e^(x)cosx - e^(x)sinx ⇒ f'(0) = 1. ∴ L(x) = f'(0)(x-0) + f(0) = x + 1. Now, f(x) ≈ L(x) ⇒ e^(x)cosx ≈ 1+x. Example: Use a linear approximation to estimate: ① ∛7 Soln: Let f(x) = ∛x, a = 8, f(8) = ∛8 = 2 and f'(x) = 1/(3x^(2/3)) ⇒ f'(8) = 1/12. L(x) = f'(8)(x-8) + f(8) = (1/12)(x-8) + 2. Now, f(x) ≈ L(x) ⇒ ∛x ≈ (1/12)(x-8) + 2. ∛7 ≈ (1/12)(-1) + 2 = 1.917 ② 256^(1/8) ≈ (1.999)+. (Exercise) ③ e^0.1 Soln: Let f(x) = e^x, a = 0. f(0) = e^0 = 1 and f'(x) = e^x ⇒ f'(0) = 1. L(x) = f'(0)(x-0) + f(0) = x + 1 ∴ e^0.1 ≈ L(0.1) ⇒ e^0.1 ≈ 0.1 + 1 = 1.1 ④ ln(1.02). (Exercise) ⑤ #28 cos 29° Soln: Let f(x) = cosx, a = π/6, f(π/6) = √3/2 and f'(x) = -sinx ⇒ f'(π/6) = -1/2. L(x) = f'(π/6)(x-π/6) + f(π/6) = -1/2(x-π/6) + √3/2. Now, f(x) ≈ L(x) ⇒ cosx ≈ -1/2(x-π/6) + √3/2. x = 29° = 29π/180 cos 29° ≈ -1/2(29π/180 - π/6) + √3/2 = π/360 + √3/2 = π/360 + 180√3/360 = (π + 180√3)/360 = 0.87475

Linear Approximations

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