Each of Exercises 15–30 gives a function f(x) and numbers L, c, and ɛ > 0. In each case, find an open interval about c on which the inequal- ity |f(x) – L| < ɛ holds. Then give a value for 8 > 0 such that for all x satisfying 0 < |x – c| < ô the inequality |f(x) – L| < ɛ holds. c = 4, L = 5, 15. f(x) — х + 1, ɛ = 0.01 16. f(x) — 2х — 2, 17. f(x) = Vx + 1, 18. f(x) = Vx, 19. f(x) —D V19 — х, L = -6, c = -2, ɛ = 0.02 c = 0, L = 1, ɛ = 0.1 L = 1/2, c = 1/4, ɛ = 0.1 L = 3, ɛ = 1 c = 10, L = 4, 20. f(x) = Vx – 7, c = 23, ɛ = 1 L = 1/4, 21. f(x) = 1/x, c = 4, ɛ = 0.05 c = V3. L = 3, 22. f(x) = x², ɛ = 0.1 %3D 23. f(x) 3D х?, 24. f(x) = 1/x, L = 4, c = -2, ɛ = 0.5 L = -1, c = -1, ɛ = 0.1 25. f(x) 3D х2 — 5, 26. f(x) —D 120/х, L = 11, c = 4, ɛ = 1 L = 5, c = 24, 27. f(x) m> 0, L= 2m, c = 2, — тх, ɛ = 0.03 28. f(x) = mx, ɛ = c > 0 L = 3m, c = 3, m > 0, L = (m/2) + b, 29. f(x) c = 1/2, m> 0, ɛ = c > 0 = mx + b, 30. f(x) 3D тх + b, m> 0, L%3Dm+ b, с %3D 1, &%3D 0.05

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.3: The Natural Exponential Function
Problem 51E
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Each of Exercises 15–30 gives a function f(x) and numbers L, c, and
ɛ > 0. In each case, find an open interval about c on which the inequal-
ity |f(x) – L| < ɛ holds. Then give a value for 8 > 0 such that for all
x satisfying 0 < |x – c| < ô the inequality |f(x) – L| < ɛ holds.
c = 4,
L = 5,
15. f(x) — х + 1,
ɛ = 0.01
16. f(x) — 2х — 2,
17. f(x) = Vx + 1,
18. f(x) = Vx,
19. f(x) —D V19 — х,
L = -6,
c = -2,
ɛ = 0.02
c = 0,
L = 1,
ɛ = 0.1
L = 1/2,
c = 1/4,
ɛ = 0.1
L = 3,
ɛ = 1
c = 10,
L = 4,
20. f(x) = Vx – 7,
c = 23,
ɛ = 1
L = 1/4,
21. f(x) = 1/x,
c = 4,
ɛ = 0.05
c = V3.
L = 3,
22. f(x) = x²,
ɛ = 0.1
%3D
23. f(x) 3D х?,
24. f(x) = 1/x,
L = 4,
c = -2,
ɛ = 0.5
L = -1,
c = -1,
ɛ = 0.1
25. f(x) 3D х2 — 5,
26. f(x) —D 120/х,
L = 11,
c = 4,
ɛ = 1
L = 5,
c = 24,
27. f(x)
m> 0, L= 2m, c = 2,
— тх,
ɛ = 0.03
28. f(x) = mx,
ɛ = c > 0
L = 3m,
c = 3,
m > 0,
L = (m/2) + b,
29. f(x)
c = 1/2,
m> 0,
ɛ = c > 0
= mx + b,
30. f(x) 3D тх + b, m> 0, L%3Dm+ b, с %3D 1, &%3D 0.05
Transcribed Image Text:Each of Exercises 15–30 gives a function f(x) and numbers L, c, and ɛ > 0. In each case, find an open interval about c on which the inequal- ity |f(x) – L| < ɛ holds. Then give a value for 8 > 0 such that for all x satisfying 0 < |x – c| < ô the inequality |f(x) – L| < ɛ holds. c = 4, L = 5, 15. f(x) — х + 1, ɛ = 0.01 16. f(x) — 2х — 2, 17. f(x) = Vx + 1, 18. f(x) = Vx, 19. f(x) —D V19 — х, L = -6, c = -2, ɛ = 0.02 c = 0, L = 1, ɛ = 0.1 L = 1/2, c = 1/4, ɛ = 0.1 L = 3, ɛ = 1 c = 10, L = 4, 20. f(x) = Vx – 7, c = 23, ɛ = 1 L = 1/4, 21. f(x) = 1/x, c = 4, ɛ = 0.05 c = V3. L = 3, 22. f(x) = x², ɛ = 0.1 %3D 23. f(x) 3D х?, 24. f(x) = 1/x, L = 4, c = -2, ɛ = 0.5 L = -1, c = -1, ɛ = 0.1 25. f(x) 3D х2 — 5, 26. f(x) —D 120/х, L = 11, c = 4, ɛ = 1 L = 5, c = 24, 27. f(x) m> 0, L= 2m, c = 2, — тх, ɛ = 0.03 28. f(x) = mx, ɛ = c > 0 L = 3m, c = 3, m > 0, L = (m/2) + b, 29. f(x) c = 1/2, m> 0, ɛ = c > 0 = mx + b, 30. f(x) 3D тх + b, m> 0, L%3Dm+ b, с %3D 1, &%3D 0.05
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