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…

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.)

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.)

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)