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)…

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.\nf(-3)= - 9 (simplify your answer.)\nnow, find f(-2).\nf(-2)= 6 (simplify your answer.)\ndoes a solution exist between - 3 and - 2 for f(x)=$pi$?\nyes, because f(-3)<$pi$<f(-2).\ninconclusive, because $pi$ does not lie between f(-3) and f(-2).\ninconclusive, because f(-3)<0 but f(-2)>0.\nyes, because f(-3)>$pi$>f(-2).\n(b) now, show that there is a value of c for which f(c)= -$sqrt{3}$. does a solution exist between - 3 and - 2?\na. inconclusive, because -$sqrt{3}$ does not lie between f(-3) and f(-2).\nb. yes, because f(-3)< -$sqrt{3}$<f(-2).\nc. yes, because f(-3)> -$sqrt{3}$>f(-2).\nd. inconclusive, because f(-3)<0 but f(-2)>0.

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.\nf(-3)= - 9 (simplify your answer.)\nnow, find f(-2).\nf(-2)= 6 (simplify your answer.)\ndoes a solution exist between - 3 and - 2 for f(x)=$pi$?\nyes, because f(-3)<$pi$<f(-2).\ninconclusive, because $pi$ does not lie between f(-3) and f(-2).\ninconclusive, because f(-3)<0 but f(-2)>0.\nyes, because f(-3)>$pi$>f(-2).\n(b) now, show that there is a value of c for which f(c)= -$sqrt{3}$. does a solution exist between - 3 and - 2?\na. inconclusive, because -$sqrt{3}$ does not lie between f(-3) and f(-2).\nb. yes, because f(-3)< -$sqrt{3}$<f(-2).\nc. yes, because f(-3)> -$sqrt{3}$>f(-2).\nd. inconclusive, because f(-3)<0 but f(-2)>0.

Answer

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))