Show that the equation x3 - 11x+2=0 has three solutions in the interval [-4, 4]. Let f(x) be equal to the left side of the equation. The function f is a polynomial, which is everywhere-continuous. Where do the solutions to the equation exist? O A. Solutions exist between values of x1 and x2 if f(x1)s0 and f(X2)50 or if f(x1)20 and f(x2)20. O B. Solutions exist at values of x1 and x2 if 0sf(x1)sf(x2) or 02f(x1)2f(x2). OC. Solutions exist at values of x1 and x2 if f(x1) f(x2). O D. Solutions exist between values of x1 and x2 if f(x1)<0 and f(x2)20 or if f(x1)20 and f(x2)<0. Find f(-4). f(-4) = (Simplify your answer.) Now, find f(- 3). f(- 3) = (Simplify your answer.) Does a solution exist between -4 and -3? O A. Yes, because 0 lies between f(- 4) and f(- 3). O B. Yes, because f(- 4)

Show that the equation x3 - 11x+2=0 has three solutions in the interval [-4, 4]. Let f(x) be equal to the left side of the equation. The function f is a polynomial, which is everywhere-continuous. Where do the solutions to the equation exist? O A. Solutions exist between values of x1 and x2 if f(x1)s0 and f(X2)50 or if f(x1)20 and f(x2)20. O B. Solutions exist at values of x1 and x2 if 0sf(x1)sf(x2) or 02f(x1)2f(x2). OC. Solutions exist at values of x1 and x2 if f(x1) f(x2). O D. Solutions exist between values of x1 and x2 if f(x1)<0 and f(x2)20 or if f(x1)20 and f(x2)<0. Find f(-4). f(-4) = (Simplify your answer.) Now, find f(- 3). f(- 3) = (Simplify your answer.) Does a solution exist between -4 and -3? O A. Yes, because 0 lies between f(- 4) and f(- 3). O B. Yes, because f(- 4)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.1: Polynomial Functions Of Degree Greater Than

Problem 49E

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Can I get with this problem step by step?

Show that the equation x3 - 11x+2=0 has three solutions in the interval [-4, 4].
Let f(x) be equal to the left side of the equation. The function f is a polynomial, which is everywhere-continuous. Where do the solutions to the equation exist?
O A. Solutions exist between values of x1 and x2 if f(x1)s0 and f(X2)50 or if f(x1)20 and f(x2)20.
O B. Solutions exist at values of x1 and x2 if 0sf(x1)sf(x2) or 02f(x1)2f(x2).
OC. Solutions exist at values of x1 and x2 if f(x1)<f(x2) or f(x1)> f(x2).
O D. Solutions exist between values of x1 and x2 if f(x1)<0 and f(x2)20 or if f(x1)20 and f(x2)<0.
Find f(-4).
f(-4) =
(Simplify your answer.)
Now, find f(- 3).
f(- 3) =
(Simplify your answer.)
Does a solution exist between -4 and -3?
O A. Yes, because 0 lies between f(- 4) and f(- 3).
O B. Yes, because f(- 4)<f(- 3).
OC. Inconclusive, because f(- 4)#0 and f(-3)#0.
O D. Inconclusive, because 0 does not lie between f(-4) and f(-3).
Now, find f(- 2).
f(- 2) =
(Simplify your answer.)
Does a solution exist between -3 and -2?
O A. Yes, because f(-3)<f(- 2).
O B. Yes, because 0 lies between f(-3) and f(- 2).
OC. Inconclusive, because 0 does not lie between f(- 3) and f(-2).
O D. Inconclusive, because f(- 3)#0 and f(-2)#0.
Continue in this manner. Where do the rest of the solutions occur? Select all that apply.
HA. Between 0 and 1
O B. Between 3 and 4
O C. Between 1 and 2
O D. Between -1 and 0
N E. Between -2 and - 1
OF. Between 2 and 3

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