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In Exercises 39–44, each function f(x) changes value when x changes from x, to xo + dx. Find a. the change Af = f(xo + dx) – f(xo); b. the value of the estimate df = f'(xo) dx; and c. the approximation error |Af – df|. y = f(x)/ Af = f(xo + dx) – f(x) df = f'(xo) dx (xo, F(xo)) dx Tangent 0| xo + dx 39. f(x) 3D х? + 2x, хо —D 1, 40. f(x) = 2x² + 4x – 3, xo = -1, dx = 0.1 41. f(x) = x³ - x, xo = 1, dx = 0.1 dx = 0.1 %3D 42. f(x) 3 х, Хо —D 1, dx %3D 0.1 43. f(x) — х 1, Хо —D 0.5, dx %3D0.1 44. f(x) 3D х3 — 2х + 3, Хо — 2, dx 3D 0.1

In Exercises 39–44, each function f(x) changes value when x changes from x, to xo + dx. Find a. the change Af = f(xo + dx) – f(xo); b. the value of the estimate df = f'(xo) dx; and c. the approximation error |Af – df|. y = f(x)/ Af = f(xo + dx) – f(x) df = f'(xo) dx (xo, F(xo)) dx Tangent 0| xo + dx 39. f(x) 3D х? + 2x, хо —D 1, 40. f(x) = 2x² + 4x – 3, xo = -1, dx = 0.1 41. f(x) = x³ - x, xo = 1, dx = 0.1 dx = 0.1 %3D 42. f(x) 3 х, Хо —D 1, dx %3D 0.1 43. f(x) — х 1, Хо —D 0.5, dx %3D0.1 44. f(x) 3D х3 — 2х + 3, Хо — 2, dx 3D 0.1

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In Exercises 39–44, each function f(x) changes value when x changes from x, to xo + dx. Find a. the change Af = f(xo + dx) – f(xo); b. the value of the estimate df = f'(xo) dx; and c. the approximation error |Af – df|. y = f(x)/ Af = f(xo + dx) – f(x) df = f'(xo) dx (xo, F(xo)) dx Tangent 0| xo + dx 39. f(x) 3D х? + 2x, хо —D 1, 40. f(x) = 2x² + 4x – 3, xo = -1, dx = 0.1 41. f(x) = x³ - x, xo = 1, dx = 0.1 dx = 0.1 %3D 42. f(x) 3 х, Хо —D 1, dx %3D 0.1 43. f(x) — х 1, Хо —D 0.5, dx %3D0.1 44. f(x) 3D х3 — 2х + 3, Хо — 2, dx 3D 0.1
Transcribed Image Text:In Exercises 39–44, each function f(x) changes value when x changes from x, to xo + dx. Find a. the change Af = f(xo + dx) – f(xo); b. the value of the estimate df = f'(xo) dx; and c. the approximation error |Af – df|. y = f(x)/ Af = f(xo + dx) – f(x) df = f'(xo) dx (xo, F(xo)) dx Tangent 0| xo + dx 39. f(x) 3D х? + 2x, хо —D 1, 40. f(x) = 2x² + 4x – 3, xo = -1, dx = 0.1 41. f(x) = x³ - x, xo = 1, dx = 0.1 dx = 0.1 %3D 42. f(x) 3 х, Хо —D 1, dx %3D 0.1 43. f(x) — х 1, Хо —D 0.5, dx %3D0.1 44. f(x) 3D х3 — 2х + 3, Хо — 2, dx 3D 0.1
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