if f(x) = x^3 - 4x + 6, show that there are values c for which f(c) equals (a) $pi$, (b) - $sqrt{3}$, and…

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. so, continue using larger values of x. find f(175). f(175)=5358681 (simplify your answer.) does a solution exist between 170 and 175 for f(x)=5,000,000? yes, because f(170)<5,000,000<f(175). inconclusive, because 5,000,000 does not lie between f(170) and f(175). inconclusive, because f(170)<0 but f(175)>0. yes, because f(170)>5,000,000>f(175).

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. so, continue using larger values of x. find f(175). f(175)=5358681 (simplify your answer.) does a solution exist between 170 and 175 for f(x)=5,000,000? yes, because f(170)<5,000,000<f(175). inconclusive, because 5,000,000 does not lie between f(170) and f(175). inconclusive, because f(170)<0 but f(175)>0. yes, because f(170)>5,000,000>f(175).

Answer

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that 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 ((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 (c)

We know (f(170)=170^{3}-4\times170 + 6=4913000-680 + 6=4912326) and (f(175)=5358681). Since (4912326<5000000<5358681), by the Intermediate - Value Theorem, there exists a number (c\in(170,175)) such that (f(c) = 5000000).

Answer:

A. Yes, because (f(170)<5000000<f(175))