Let p(x) be a cubic polynomial with p(7) < 0,p(12) > 0 and p(14) < 0. What can you say about the number and location of zeros p(x)? O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14. p(x) has a repeated zero. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 andx= 14. p(x) has a third zero greater than 14 or less than 7. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 andx= 14. p(x) has a third zero greater than 14 or less than 12. O p(x) has one zero between x = 7 and x = 14. p(x) has a repeated zero. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14. p(x) has a third zero greater than 12 or less than 7.
Let p(x) be a cubic polynomial with p(7) < 0,p(12) > 0 and p(14) < 0. What can you say about the number and location of zeros p(x)? O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14. p(x) has a repeated zero. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 andx= 14. p(x) has a third zero greater than 14 or less than 7. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 andx= 14. p(x) has a third zero greater than 14 or less than 12. O p(x) has one zero between x = 7 and x = 14. p(x) has a repeated zero. O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14. p(x) has a third zero greater than 12 or less than 7.
Calculus: Early Transcendentals
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Author:James Stewart
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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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![Show that there is a number c, with 0 < c < 1, such that f(c) = 0.
f(x) = 2x- cos x
%3D
O We have that f(0) = 1 > 0 and f(1) = 2 – cos 1 > 0 and thatf is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = – 1 < 0 and f(1) = 2 – cos 1 > 0 and thatf is continuous. Thus, by the Intermediate Value
Theorem applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f (0) = – 1 < 0 and f(1) = 2 – cos 1 > 0 and thatf is periodic. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = – 1 < 0 and f(1) = 2 > 0 and thatf is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = 1 > 0 and f(1) = 2 > 0 and that f is periodic. Thus, by the Intermediate Value Theorem applied to
k = 0, there is a number c in [0, 1] such that f(c) = k = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1942d21f-1f68-402e-935c-586383b38458%2F1cac7926-a1c7-4cc9-adb9-f7f98d7747e7%2Fnw41zq1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show that there is a number c, with 0 < c < 1, such that f(c) = 0.
f(x) = 2x- cos x
%3D
O We have that f(0) = 1 > 0 and f(1) = 2 – cos 1 > 0 and thatf is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = – 1 < 0 and f(1) = 2 – cos 1 > 0 and thatf is continuous. Thus, by the Intermediate Value
Theorem applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f (0) = – 1 < 0 and f(1) = 2 – cos 1 > 0 and thatf is periodic. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = – 1 < 0 and f(1) = 2 > 0 and thatf is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
O We have that f(0) = 1 > 0 and f(1) = 2 > 0 and that f is periodic. Thus, by the Intermediate Value Theorem applied to
k = 0, there is a number c in [0, 1] such that f(c) = k = 0.
Transcribed Image Text:Let p(x) be a cubic polynomial with p(7) < 0, p(12) > 0 and p(14) < 0. What can you say about the number and location of zeros
p(x)?
O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14.
p(x) has a repeated zero.
O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14.
p(x) has a third zero greater than 14 or less than 7.
O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 andx = 14.
p(x) has a third zero greater than 14 or less than 12.
O p(x) has one zero between x = 7 and = 14.
p(x) has a repeated zero.
O p(x) has a zero between x = 7 and x = 12 and another betweenx = 12 and x = 14.
p(x) has a third zero greater than 12 or less than 7.
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