if f(x)=x^3 - 4x + 6, show that there are values c for which f(c) equals (a) π, (b) -√3, and (c) 5,000,000…

if f(x)=x^3 - 4x + 6, show that there are values c for which f(c) equals (a) π, (b) -√3, and (c) 5,000,000. a. find values of x1 and x2 for which f(x1)≤π≤f(x2) or f(x1)≥π≥f(x2). b. find values of x1 and x2 for which f(x1)≤0≤f(x2) or f(x1)≥0≥f(x2). c. find values of x1 and x2 for which f(x1)=0 or f(x2)=0. d. find values of x1 and x2 for which f(x1)≤π and f(x2)≤π or f(x1)≥π and f(x2)≥π. use the trial and error method. begin by finding f(-3). f(-3)= -9 (simplify your answer.) now, find f(-2). f(-2)= 6 (simplify your answer.) does a solution exist between -3 and -2 for f(x)=π? yes, because f(-3)<π<f(-2). inconclusive, because π does not lie between f(-3) and f(-2). inconclusive, because f(-3)<0 but f(-2)>0. yes, because f(-3)>π>f(-2).

if f(x)=x^3 - 4x + 6, show that there are values c for which f(c) equals (a) π, (b) -√3, and (c) 5,000,000. a. find values of x1 and x2 for which f(x1)≤π≤f(x2) or f(x1)≥π≥f(x2). b. find values of x1 and x2 for which f(x1)≤0≤f(x2) or f(x1)≥0≥f(x2). c. find values of x1 and x2 for which f(x1)=0 or f(x2)=0. d. find values of x1 and x2 for which f(x1)≤π and f(x2)≤π or f(x1)≥π and f(x2)≥π. use the trial and error method. begin by finding f(-3). f(-3)= -9 (simplify your answer.) now, find f(-2). f(-2)= 6 (simplify your answer.) does a solution exist between -3 and -2 for f(x)=π? yes, because f(-3)<π<f(-2). inconclusive, because π does not lie between f(-3) and f(-2). inconclusive, because f(-3)<0 but f(-2)>0. yes, because f(-3)>π>f(-2).

Answer

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function (y = f(x)) is continuous on a closed interval ([a,b]), and (k) is a number between (f(a)) and (f(b)), then there exists at least one number (c) in the interval ((a,b)) such that (f(c)=k). For showing (f(c)=\pi), we need to find (x_1) and (x_2) such that (f(x_1)\leq\pi\leq f(x_2)) or (f(x_1)\geq\pi\geq f(x_2)).

Step2: Analyze given values

We are given (f(-3)=-9) and (f(-2) = 6). Since (-9=\ f(-3)<\pi\approx3.14<f(-2) = 6), by the Intermediate - Value Theorem, there exists a (c\in(-3,-2)) such that (f(c)=\pi).

Answer:

A. Find values of (x_1) and (x_2) for which (f(x_1)\leq\pi\leq f(x_2)) or (f(x_1)\geq\pi\geq f(x_2)) Yes, because (f(-3)<\pi<f(-2))