Question

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).

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 \(f(-3)<\pi<f(-2)\)