Show that the equation x³ – 3x – 1 has three solutions in the interval [-3, 3]. 1. Let f(x) be equal to the left side of the equation. In which situation are we guaranteed that solutions to the equation exist? (a) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2) > 0, or f(x1) > 0 and f(x2) < 0. (b) Solutions are guaranteed to exist at values of x1 and x2 if 0 < f(x1) < f(x2) or 0 < f(x2)< f(x1). (c) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2)< 0, or f(x1) > 0 and f(x2) > 0. (d) Solutions are guaranteed to exist at values of x1 and x2 if f(x1) < f(x2) or f(x2) < f(x1). 2. Does a solution exist between 1 and 2? (a) Yes, because f(1) < f(2). (b) Yes, because 0 lies between f(1) and f(2). (c) Inconclusive, because 0 does not lie between f(1) and f(2). (d) Inconclusive, because f(1) 0 and f(2) 0. 3. Does a solution exist between 0 and 1? (a) Yes, because f(0) < f(1). (b) Yes, because 0 lies between f(1) and f(0). (c) Inconclusive, because 0 does not lie between f(1) and f(0). (d) Inconclusive, because f(1) + 0 and f(0) 0. 4. Where do the rest of the solutions occur? Select all that apply. (a) Between 2 and 3. (b) Between -3 and -2. (d) Between –1 and 0. (c) Between -2 and –1.
Show that the equation x³ – 3x – 1 has three solutions in the interval [-3, 3]. 1. Let f(x) be equal to the left side of the equation. In which situation are we guaranteed that solutions to the equation exist? (a) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2) > 0, or f(x1) > 0 and f(x2) < 0. (b) Solutions are guaranteed to exist at values of x1 and x2 if 0 < f(x1) < f(x2) or 0 < f(x2)< f(x1). (c) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2)< 0, or f(x1) > 0 and f(x2) > 0. (d) Solutions are guaranteed to exist at values of x1 and x2 if f(x1) < f(x2) or f(x2) < f(x1). 2. Does a solution exist between 1 and 2? (a) Yes, because f(1) < f(2). (b) Yes, because 0 lies between f(1) and f(2). (c) Inconclusive, because 0 does not lie between f(1) and f(2). (d) Inconclusive, because f(1) 0 and f(2) 0. 3. Does a solution exist between 0 and 1? (a) Yes, because f(0) < f(1). (b) Yes, because 0 lies between f(1) and f(0). (c) Inconclusive, because 0 does not lie between f(1) and f(0). (d) Inconclusive, because f(1) + 0 and f(0) 0. 4. Where do the rest of the solutions occur? Select all that apply. (a) Between 2 and 3. (b) Between -3 and -2. (d) Between –1 and 0. (c) Between -2 and –1.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 15E
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![Show that the equation x³ – 3x – 1 has three solutions in the interval [-3, 3].
1. Let f(x) be equal to the left side of the equation. In which situation are we guaranteed that
solutions to the equation exist?
(a) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2) > 0,
or f(x1) > 0 and f(x2) < 0.
(b) Solutions are guaranteed to exist at values of x1 and x2 if 0 < f(x1) < f(x2) or 0 < f(x2)<
f(x1).
(c) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2)< 0,
or f(x1) > 0 and f(x2) > 0.
(d) Solutions are guaranteed to exist at values of x1 and x2 if f(x1) < f(x2) or f(x2) < f(x1).
2. Does a solution exist between 1 and 2?
(a) Yes, because f(1) < f(2).
(b) Yes, because 0 lies between f(1) and f(2).
(c) Inconclusive, because 0 does not lie between f(1) and f(2).
(d) Inconclusive, because f(1) 0 and f(2) 0.
3. Does a solution exist between 0 and 1?
(a) Yes, because f(0) < f(1).
(b) Yes, because 0 lies between f(1) and f(0).
(c) Inconclusive, because 0 does not lie between f(1) and f(0).
(d) Inconclusive, because f(1) + 0 and f(0) 0.
4. Where do the rest of the solutions occur? Select all that apply.
(a) Between 2 and 3.
(b) Between -3 and -2.
(d) Between –1 and 0.
(c) Between -2 and –1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc31d976c-acd4-451e-8eb9-fde305025b60%2F9b85f515-c341-42d0-bf9b-aff453356e73%2Fu3g7l1g_processed.png&w=3840&q=75)
Transcribed Image Text:Show that the equation x³ – 3x – 1 has three solutions in the interval [-3, 3].
1. Let f(x) be equal to the left side of the equation. In which situation are we guaranteed that
solutions to the equation exist?
(a) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2) > 0,
or f(x1) > 0 and f(x2) < 0.
(b) Solutions are guaranteed to exist at values of x1 and x2 if 0 < f(x1) < f(x2) or 0 < f(x2)<
f(x1).
(c) Solutions are guaranteed to exist between the values of x1 and x2 if f(x1) < 0 and f(x2)< 0,
or f(x1) > 0 and f(x2) > 0.
(d) Solutions are guaranteed to exist at values of x1 and x2 if f(x1) < f(x2) or f(x2) < f(x1).
2. Does a solution exist between 1 and 2?
(a) Yes, because f(1) < f(2).
(b) Yes, because 0 lies between f(1) and f(2).
(c) Inconclusive, because 0 does not lie between f(1) and f(2).
(d) Inconclusive, because f(1) 0 and f(2) 0.
3. Does a solution exist between 0 and 1?
(a) Yes, because f(0) < f(1).
(b) Yes, because 0 lies between f(1) and f(0).
(c) Inconclusive, because 0 does not lie between f(1) and f(0).
(d) Inconclusive, because f(1) + 0 and f(0) 0.
4. Where do the rest of the solutions occur? Select all that apply.
(a) Between 2 and 3.
(b) Between -3 and -2.
(d) Between –1 and 0.
(c) Between -2 and –1.
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