If f(x)=x² - 6x +8, show that there are values c for which f(c) equals (a), (b) -√3, and (c) 6,000,000. (a) The function f is a polynomial, which is everywhere-continuous. What is the best way to begin the problem? OA. Find values of x, and x₂ for which f(x₁) s0s f(x₂) or f(x₁) 20≥f(x₂). OB. Find values of x, and x₂ for which f(x₁) = 0 or f(x₂) = 0. OC. Find values of x, and x₂ for which f(x₁) sx and f(x₂) sx or f(x₁) 2x and f(x₂) 2*. O D. Find values of x, and x₂ for which f(x₁) sas f(x₂) or f(x₁)2x21(x₂). Use the trial and error method. Begin by finding f(-4). f(-4)= (Simplify your answer.) Now, find f(-2). f(-2) = (Simplify your answer.) Does a solution exist between -4 and -2 for f(x)=x? O Inconclusive, because x does not lie between f(-4) and f(-2). O Yes, because f(-4) >x>f(-2). O Inconclusive, because f(-4) <0 but f(-2) >0. O Yes, because f(-4)
If f(x)=x² - 6x +8, show that there are values c for which f(c) equals (a), (b) -√3, and (c) 6,000,000. (a) The function f is a polynomial, which is everywhere-continuous. What is the best way to begin the problem? OA. Find values of x, and x₂ for which f(x₁) s0s f(x₂) or f(x₁) 20≥f(x₂). OB. Find values of x, and x₂ for which f(x₁) = 0 or f(x₂) = 0. OC. Find values of x, and x₂ for which f(x₁) sx and f(x₂) sx or f(x₁) 2x and f(x₂) 2*. O D. Find values of x, and x₂ for which f(x₁) sas f(x₂) or f(x₁)2x21(x₂). Use the trial and error method. Begin by finding f(-4). f(-4)= (Simplify your answer.) Now, find f(-2). f(-2) = (Simplify your answer.) Does a solution exist between -4 and -2 for f(x)=x? O Inconclusive, because x does not lie between f(-4) and f(-2). O Yes, because f(-4) >x>f(-2). O Inconclusive, because f(-4) <0 but f(-2) >0. O Yes, because f(-4)
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Transcribed Image Text:If f(x)=x² - 6x+8, show that there are values c for which f(c) equals (a), (b) -√3, and (c) 6,000,000.
(a) The function f is a polynomial, which is everywhere-continuous. What is the best way to begin the problem?
O A. Find values of x, and x₂ for which f(x₁) ≤0≤ f(x₂) or f(x₁) 20≥ f(x₂).
OB. Find values of x, and x₂ for which f(x₁) = 0 or f(x₂) = 0.
OC. Find values of x₁ and x₂ for which f(x₁) ≤ and f(x₂) S or f(x₁) and f(x₂) ².
OD. Find values of x, and x₂ for which f(x₁) ss f(x₂) or f(x₁)x21(x₂).
Use the trial and error method. Begin by finding f(-4).
f(-4)= (Simplify your answer.)
Now, find f(-2).
f(-2)=(Simplify your answer.)
Does a solution exist between 4 and -2 for f(x)=x?
Inconclusive, because x does not lie between f(-4) and f(-2).
OYes, because f(-4) >x>f(-2).
O Inconclusive, because f(-4) <0 but f(-2) >0.
OYes, because f(-4)<x<f(-2).
Transcribed Image Text:(b) Now, show that there is a value of c for which f(c)= -√3. Does a solution exist between -4 and -27
OA. Inconclusive, because -√3 does not lie between f(-4) and f(-2).
OB. Inconclusive, because f(-4) <0 but f(-2) >0.
OC. Yes, because f(-4)<-√3<f(-2).
OD. Yes, because f(-4) > -√3>1(-2).
(c) Finally, show that there is a value of c for which f(c)=6,000,000. Start with a large x-value. In this case, use 175. Find f(175).
f(175)=
(Simplify your answer.)
Now, find 1(180).
f(180) = (Simplify your answer.)
Does a solution exist between 175 and 180 for f(x)=6,000,000?
Inconclusive, because f(175) <0 but f(180) > 0.
Yes, because f(175) <6,000,000<f(180).
Inconclusive, because 6,000,000 does not lie between f(175) and f(180).
Yes, because f(175) >6,000,000>(180).
So, continue using larger values of x. Find f(185).
((185)= (Simplify your answer.)
Does a solution exist between 180 and 185 for f(x)=6,000,000?
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