if f(x)=x³ - 4x + 6, show that there are values c for which f(c) equals (a) π, (b) -√3, and (c)…

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.\nf(-2)=6 (simplify your answer.)\ndoes a solution exist between -3 and -2 for f(x)=π?\nyes, because f(-3)<π<f(-2).\ninconclusive, because π does not lie between f(-3) and f(-2).\ninconclusive, because f(-3)<0 but f(-2)>0.\nyes, because f(-3)>π>f(-2).\n(b) now, show that there is a value of c for which f(c)= -√3. does a solution exist between -3 and -2?\na. inconclusive, because -√3 does not lie between f(-3) and f(-2).\nb. yes, because f(-3)< -√3<f(-2).\nc. yes, because f(-3)> -√3>f(-2).\nd. inconclusive, because f(-3)<0 but f(-2)>0.\n(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).\nf(165)= (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.\nf(-2)=6 (simplify your answer.)\ndoes a solution exist between -3 and -2 for f(x)=π?\nyes, because f(-3)<π<f(-2).\ninconclusive, because π does not lie between f(-3) and f(-2).\ninconclusive, because f(-3)<0 but f(-2)>0.\nyes, because f(-3)>π>f(-2).\n(b) now, show that there is a value of c for which f(c)= -√3. does a solution exist between -3 and -2?\na. inconclusive, because -√3 does not lie between f(-3) and f(-2).\nb. yes, because f(-3)< -√3<f(-2).\nc. yes, because f(-3)> -√3>f(-2).\nd. inconclusive, because f(-3)<0 but f(-2)>0.\n(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).\nf(165)= (simplify your answer.)

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)