Question

if f(x)=x^3 - 4x + 6, show that there are values c for which f(c) equals (a) $pi$, (b) -$sqrt{3}$, and (c) 5,000,000.
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)=$pi$?
yes, because f(-3)<$pi$<f(-2).
inconclusive, because $pi$ does not lie between f(-3) and f(-2).
inconclusive, because f(-3)<0 but f(-2)>0.
yes, because f(-3)>$pi$>f(-2).
(b) now, show that there is a value of c for which f(c)= -$sqrt{3}$. does a solution exist between - 3 and - 2?
a. inconclusive, because -$sqrt{3}$ does not lie between f(-3) and f(-2).
b. yes, because f(-3)< -$sqrt{3}$<f(-2).
c. yes, because f(-3)> -$sqrt{3}$>f(-2).
d. inconclusive, because f(-3)<0 but f(-2)>0.

Explanation:

Step1: Recall Intermediate - Value Theorem

If \(y = f(x)\) 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 \(f(c)=k\). The function \(f(x)=x^{3}-4x + 6\) is a polynomial, so it is continuous everywhere.

Step2: Analyze part (a)

We know \(f(-3)=-9\) and \(f(-2)=6\). Since \(-9<\pi<6\), by the Intermediate - Value Theorem, there exists a \(c\in(-3,-2)\) such that \(f(c)=\pi\).

Step3: Analyze part (b)

We know \(f(-3)=-9\) and \(f(-2)=6\). Also, \(-9<-\sqrt{3}<6\). Since \(-\sqrt{3}\approx - 1.732\), and \(f(x)\) is continuous on \([-3,-2]\), by the Intermediate - Value Theorem, there exists a \(c\in(-3,-2)\) such that \(f(c)=-\sqrt{3}\).

Answer:

For the first question (a): Yes, because \(f(-3)<\pi<f(-2)\)
For the second question (b): B. Yes, because \(f(-3)<-\sqrt{3}<f(-2)\)