Question

if f(x)=x³ - 4x + 6, show that there are values c for which f(c) equals (a) π, (b) -√3, and (c) 5,000,000. (a) the function f is a polynomial, which is everywhere - continuous. what is the best way to begin the problem? 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.

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 a continuous function \(f(x)\) and a target value \(k\), we need to find \(x_1\) and \(x_2\) such that \(k\) lies between \(f(x_1)\) and \(f(x_2)\). Here \(k = \pi\), so we need to find \(x_1\) and \(x_2\) for which \(f(x_1)\leq\pi\leq f(x_2)\) or \(f(x_1)\geq\pi\geq f(x_2)\).

Answer:

A. Find values of \(x_1\) and \(x_2\) for which \(f(x_1)\leq\pi\leq f(x_2)\) or \(f(x_1)\geq\pi\geq f(x_2)\).