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.
f(-2)=6 (simplify your answer.)
does a solution exist between -3 and -2 for f(x)=π?
yes, because f(-3)<π<f(-2).
inconclusive, because π does not lie between f(-3) and f(-2).
inconclusive, because f(-3)<0 but f(-2)>0.
yes, because f(-3)>π>f(-2).
(b) now, show that there is a value of c for which f(c)= -√3. does a solution exist between -3 and -2?
a. inconclusive, because -√3 does not lie between f(-3) and f(-2).
b. yes, because f(-3)< -√3<f(-2).
c. yes, because f(-3)> -√3>f(-2).
d. inconclusive, because f(-3)<0 but f(-2)>0.
(c) finally, show that there is a value of c for which f(c)=5,000,000. start with a large x - value. in this case, use 165. find f(165).
f(165)= (simplify your answer.)

Explanation:

Step1: Calculate \(f(-3)\)

Substitute \(x = - 3\) into \(f(x)=x^{3}-4x + 6\). So \(f(-3)=(-3)^{3}-4\times(-3)+6=-27 + 12+6=-9\).

Step2: Analyze part (a)

We know \(f(-3)=-9\), \(f(-2) = 6\) and \(\pi\approx3.14\). Since \(-9<\pi<6\), by the Intermediate - Value Theorem (if \(y = f(x)\) is continuous on the closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \((a,b)\) such that \(f(c)=k\)), for \(a=-3\), \(b = - 2\) and \(k=\pi\), a solution exists between \(-3\) and \(-2\) because \(f(-3)<\pi<f(-2)\).

Step3: Analyze part (b)

We know \(f(-3)=-9\), \(f(-2)=6\) and \(-\sqrt{3}\approx - 1.73\). Since \(-9<-\sqrt{3}<6\), by the Intermediate - Value Theorem, for \(a=-3\), \(b=-2\) and \(k =-\sqrt{3}\), a solution exists between \(-3\) and \(-2\) because \(f(-3)<-\sqrt{3}<f(-2)\).

Step4: Calculate \(f(165)\) for part (c)

Substitute \(x = 165\) into \(f(x)=x^{3}-4x + 6\).
\[

$$\begin{align*} f(165)&=165^{3}-4\times165 + 6\\ &=165\times165\times165-660 + 6\\ &=4492125-660+6\\ &=4491471 \end{align*}$$

\]
Since \(f(x)=x^{3}-4x + 6\) is a polynomial function, it is continuous everywhere. As \(x\) increases, \(f(x)\) also increases. \(f(165)=4491471<5000000\). But as \(x\) gets larger, \(f(x)\) will exceed \(5000000\).

Answer:

(a) Yes, because \(f(-3)<\pi<f(-2)\)
(b) Yes, because \(f(-3)<-\sqrt{3}<f(-2)\)
(c) \(f(165)=4491471\)