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. use the trial and error method. begin by finding f(-3). f(-3)=□ (simplify your answer.)

Explanation:

Step1: Recall the Intermediate - Value Theorem

For a continuous function \(y = f(x)\) on an interval \([a,b]\), if \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \((a,b)\) such that \(f(c)=k\). Here, we want to find \(x_1\) and \(x_2\) such that \(f(x_1)\leq\pi\leq f(x_2)\) or \(f(x_1)\geq\pi\geq f(x_2)\) to show the existence of \(c\) with \(f(c)=\pi\). So the best way to begin is option A.

Step2: Calculate \(f(-3)\)

Substitute \(x=-3\) into \(f(x)=x^{3}-4x + 6\).
\[

$$\begin{align*} f(-3)&=(-3)^{3}-4\times(-3)+6\\ &=-27 + 12+6\\ &=-9 \end{align*}$$

\]

Answer:

A. Find values of \(x_1\) and \(x_2\) for which \(f(x_1)\leq\pi\leq f(x_2)\) or \(f(x_1)\geq\pi\geq f(x_2)\); \(f(-3)=-9\)