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)= - 9 (simplify your answer.)
now, find f(-2).
f(-2)= (simplify your answer.)

Explanation:

Step1: Recall polynomial property

The polynomial function $f(x)=x^{3}-4x + 6$ is continuous everywhere. By the Intermediate - Value Theorem, if we can find two points $x_1$ and $x_2$ such that $f(x_1)\leq k\leq f(x_2)$ or $f(x_1)\geq k\geq f(x_2)$ for a given value $k$, then there exists a $c$ in the interval $(x_1,x_2)$ such that $f(c)=k$. 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 $f(x_1)\geq\pi\geq f(x_2)$.

Step2: Calculate $f(-2)$

Substitute $x=-2$ into $f(x)=x^{3}-4x + 6$.
\[

$$\begin{align*} f(-2)&=(-2)^{3}-4\times(-2)+6\\ &=-8 + 8+6\\ &=6 \end{align*}$$

\]

Answer:

6