1. The linearization at a = 0 to v8+ &x is A + Bx. Compute A and B. %3D A = 2. The linearization at a = 0 to sin(3x) is A + Bx. Compute A and B. A = B 3. Find the linearization L(x) of the function g(x) =xf(x²) at x = 2 given the following infor-mation. f (2) = -1f'(2) = 12f (4) = 6 f'(4) = -2 Answer: L(x) =. 4. The figure below shows f(x) and its local linearization at x = a, y = 3x – 1. (The local linearization is shown in blue.) What is the value of a? a = What is the value of f(a)? f(a) =- Use the linearization to approximate the value of f(2.5). f(2.5) = Is the approximation an under- or overestimate? (Enter under or over.) 5. The figure shows how a function f(x) and its linear approximation (i.e., its tangent line) change value when x changes from xo to xo+dx. Suppose f(x) = x² – 3x, xo = 3 and dx = 0.05. Your answers below need to be very precise, so use many decimal places. (a) Find the change Af = f(xo+dx) – f(xo). Af =, (b) Find the estimate (i.e., the differential) df = f'(xo) dx. df = . (c) Find the approximation error |Af –df\. Error =. y = f(x) f(xo+ dx) Error = |Af - df| Af = f(x0+ dx) – f(xo) Tangent |df = f'(x)dx f(xo) dx xo+ dx (Click on graph to enlarge)
1. The linearization at a = 0 to v8+ &x is A + Bx. Compute A and B. %3D A = 2. The linearization at a = 0 to sin(3x) is A + Bx. Compute A and B. A = B 3. Find the linearization L(x) of the function g(x) =xf(x²) at x = 2 given the following infor-mation. f (2) = -1f'(2) = 12f (4) = 6 f'(4) = -2 Answer: L(x) =. 4. The figure below shows f(x) and its local linearization at x = a, y = 3x – 1. (The local linearization is shown in blue.) What is the value of a? a = What is the value of f(a)? f(a) =- Use the linearization to approximate the value of f(2.5). f(2.5) = Is the approximation an under- or overestimate? (Enter under or over.) 5. The figure shows how a function f(x) and its linear approximation (i.e., its tangent line) change value when x changes from xo to xo+dx. Suppose f(x) = x² – 3x, xo = 3 and dx = 0.05. Your answers below need to be very precise, so use many decimal places. (a) Find the change Af = f(xo+dx) – f(xo). Af =, (b) Find the estimate (i.e., the differential) df = f'(xo) dx. df = . (c) Find the approximation error |Af –df\. Error =. y = f(x) f(xo+ dx) Error = |Af - df| Af = f(x0+ dx) – f(xo) Tangent |df = f'(x)dx f(xo) dx xo+ dx (Click on graph to enlarge)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 93E
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