Show that the equation x - 14x+ 6 = 0 has three solutions in the interval [- 4, 4]. Let f(x) be equal to the left side of the equation. Where do the solutions to the equation exist? O A. Solutions exist at values of x, and x, if f(x,)< f(x,) or f(x,) > f(x). O B. Solutions exist between values of x, and x, if f(x,)s0 and f(x,) 20 or f(x,)20 and f(x2)s0. OC. Solutions exist between values of x, and x, if f(x,)s0 and f(x2)s0 or f(x,)20 and f(x2) 20. O D. Solutions exist at values of x, and x, if 0s f(x,)sf(x2) or 0 2 f(x,)2 f(x2). 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. Inconclusive, because f(- 4) #0 and f( - 3)#0. O B. Inconclusive, because 0 does not lie between f(- 4) and f( - 3). OC. Yes, because f( - 4) < f(- 3). O D. Yes, because 0 lies between f(- 4) and f( - 3). Now find f( - 2). f(- 2) =D (Simplify your answer.) Does a solution exist between -3 and - 2? O Yes, because O lies between f( - 3) and f( - 2). O Inconclusive, because 0 does not lie between f(- 3) and f( - 2). O Yes, because f( - 3) < f(- 2). O 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. OA. Between 2 and 3 O B. Between 1 and 2 O C. Between 0 and 1 O D. Between - 2 and - 1 O E. Between - 1 and 0 OF. Between 3 and 4
Show that the equation x - 14x+ 6 = 0 has three solutions in the interval [- 4, 4]. Let f(x) be equal to the left side of the equation. Where do the solutions to the equation exist? O A. Solutions exist at values of x, and x, if f(x,)< f(x,) or f(x,) > f(x). O B. Solutions exist between values of x, and x, if f(x,)s0 and f(x,) 20 or f(x,)20 and f(x2)s0. OC. Solutions exist between values of x, and x, if f(x,)s0 and f(x2)s0 or f(x,)20 and f(x2) 20. O D. Solutions exist at values of x, and x, if 0s f(x,)sf(x2) or 0 2 f(x,)2 f(x2). 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. Inconclusive, because f(- 4) #0 and f( - 3)#0. O B. Inconclusive, because 0 does not lie between f(- 4) and f( - 3). OC. Yes, because f( - 4) < f(- 3). O D. Yes, because 0 lies between f(- 4) and f( - 3). Now find f( - 2). f(- 2) =D (Simplify your answer.) Does a solution exist between -3 and - 2? O Yes, because O lies between f( - 3) and f( - 2). O Inconclusive, because 0 does not lie between f(- 3) and f( - 2). O Yes, because f( - 3) < f(- 2). O 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. OA. Between 2 and 3 O B. Between 1 and 2 O C. Between 0 and 1 O D. Between - 2 and - 1 O E. Between - 1 and 0 OF. Between 3 and 4
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter9: Systems Of Linear Equations
Section9.7: Puzzle Problems
Problem 8OE
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![Show that the equation x - 14x + 6= 0 has three solutions in the interval [-4, 4].
Let f(x) be equal to the left side of the equation. Where do the solutions to the equation exist?
O A. Solutions exist at values of x, and x, if f(x,)<f(x2) or f(x,) > f(x2).
O B. Solutions exist between values of x, and x, if f(x,)s0 and f(x,) 20 or f(x,)20 and f(x,)<0.
OC. Solutions exist between values of x, and x, if f(x,)s 0 and f(x,)s0 or f(x,)2 0 and f(x,) 20.
O D. Solutions exist at values of x, and x, if 0s f(x,)s f(x2) or 0 2 f(x, )2 f(x2).
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. Inconclusive, because f(- 4) # 0 and f(- 3) +0.
O B. Inconclusive, because 0 does not lie between f( - 4) and f( - 3).
OC. Yes, because f( - 4) < f(- 3).
O D. Yes, because 0 lies between f( - 4) and f( - 3).
Now find f(- 2).
f(- 2) = (Simplify your answer.)
Does a solution exist between - 3 and - 2?
Yes, because 0 lies between f( - 3) and f( - 2).
Inconclusive, because 0 does not lie between f( – 3) and f( - 2).
Yes, because f( - 3) < f(- 2).
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.
DA. Between 2 and 3
O B. Between 1 and 2
O C. Between 0 and 1
O D. Between -2 and - 1
O E. Between - 1 and 0
O F. Between 3 and 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f7775f3-edad-4f27-ba1b-f7886a7505bd%2F6a7311c9-96ed-4d70-83df-9adfd90bf136%2Fxwmudqj_processed.png&w=3840&q=75)
Transcribed Image Text:Show that the equation x - 14x + 6= 0 has three solutions in the interval [-4, 4].
Let f(x) be equal to the left side of the equation. Where do the solutions to the equation exist?
O A. Solutions exist at values of x, and x, if f(x,)<f(x2) or f(x,) > f(x2).
O B. Solutions exist between values of x, and x, if f(x,)s0 and f(x,) 20 or f(x,)20 and f(x,)<0.
OC. Solutions exist between values of x, and x, if f(x,)s 0 and f(x,)s0 or f(x,)2 0 and f(x,) 20.
O D. Solutions exist at values of x, and x, if 0s f(x,)s f(x2) or 0 2 f(x, )2 f(x2).
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. Inconclusive, because f(- 4) # 0 and f(- 3) +0.
O B. Inconclusive, because 0 does not lie between f( - 4) and f( - 3).
OC. Yes, because f( - 4) < f(- 3).
O D. Yes, because 0 lies between f( - 4) and f( - 3).
Now find f(- 2).
f(- 2) = (Simplify your answer.)
Does a solution exist between - 3 and - 2?
Yes, because 0 lies between f( - 3) and f( - 2).
Inconclusive, because 0 does not lie between f( – 3) and f( - 2).
Yes, because f( - 3) < f(- 2).
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.
DA. Between 2 and 3
O B. Between 1 and 2
O C. Between 0 and 1
O D. Between -2 and - 1
O E. Between - 1 and 0
O F. Between 3 and 4
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