8. (10 points) answer each question below.\n(a) (2 points) select an answer choice that correctly completes…

8. (10 points) answer each question below.\n(a) (2 points) select an answer choice that correctly completes the statement:\nthe intermediate value theorem for polynomials states that if ( p(x) ) is a polynomial\nand ( p(a) ) and ( p(b) ) have opposite signs, then;\na. ( p(x) ) has no real zero between ( x = a ) and ( x = b )\nb. there is at least one number ( c ) between ( a ) and ( b ) such that ( p(c)=0 )\nc. ( p(x) ) must be increasing on the interval ( (a, b) )\nd. ( p(x) ) has exactly one real zero between ( x = a ) and ( x = b )\n(b) (4 points) let ( f(x)=x^{3}+3 x - 5 ). find ( f(1) ) and ( f(2) ).\nwrite your answer in the box below:\n( f(1)= )( f(2)= )\n(c) (4 points) using your answer from (b), why does the intermediate value theorem\nfor polynomials guarantee that ( f(x)=0 ) has a solution in ( 1,2 )?\na. ( f(x) ) is a polynomial and ( f(1) ) and ( f(2) ) have opposite signs so ( f(x) ) has\nat least one ( c ) in the interval ( (1,2) ) where ( f(c)=0 )\nb. the intermediate value theorem cannot be applied here because ( f(x) ) is\nnot a polynomial\nc. ( f(x) ) is a polynomial, but ( f(1) ) and ( f(2) ) have the same sign, so there is\nno ( c ) in ( (1,2) ) such that ( f(c)=0 )

8. (10 points) answer each question below.\n(a) (2 points) select an answer choice that correctly completes the statement:\nthe intermediate value theorem for polynomials states that if ( p(x) ) is a polynomial\nand ( p(a) ) and ( p(b) ) have opposite signs, then;\na. ( p(x) ) has no real zero between ( x = a ) and ( x = b )\nb. there is at least one number ( c ) between ( a ) and ( b ) such that ( p(c)=0 )\nc. ( p(x) ) must be increasing on the interval ( (a, b) )\nd. ( p(x) ) has exactly one real zero between ( x = a ) and ( x = b )\n(b) (4 points) let ( f(x)=x^{3}+3 x - 5 ). find ( f(1) ) and ( f(2) ).\nwrite your answer in the box below:\n( f(1)= )( f(2)= )\n(c) (4 points) using your answer from (b), why does the intermediate value theorem\nfor polynomials guarantee that ( f(x)=0 ) has a solution in ( 1,2 )?\na. ( f(x) ) is a polynomial and ( f(1) ) and ( f(2) ) have opposite signs so ( f(x) ) has\nat least one ( c ) in the interval ( (1,2) ) where ( f(c)=0 )\nb. the intermediate value theorem cannot be applied here because ( f(x) ) is\nnot a polynomial\nc. ( f(x) ) is a polynomial, but ( f(1) ) and ( f(2) ) have the same sign, so there is\nno ( c ) in ( (1,2) ) such that ( f(c)=0 )

Answer

(a)

Brief Explanations:

The Intermediate Value Theorem for Polynomials states that if (P(x)) is a polynomial and (P(a)) and (P(b)) have opposite signs, then there is at least one number (c) between (a) and (b) such that (P(c)=0). Option A is incorrect because it contradicts the theorem. Option C is wrong as the theorem does not state anything about the function being increasing. Option D is incorrect because it says "exactly one" - the theorem only guarantees at least one.

Answer:

B. There is at least one number (c) between (a) and (b) such that (P(c) = 0)

(b)

Explanation:

Step1: Calculate (f(1))

Substitute (x = 1) into (f(x)=x^{3}+3x - 5) (f(1)=1^{3}+3\times1 - 5) (=1 + 3-5) (=-1)

Step2: Calculate (f(2))

Substitute (x = 2) into (f(x)=x^{3}+3x - 5) (f(2)=2^{3}+3\times2 - 5) (=8 + 6-5) (=9)

Answer:

(f(1)=-1), (f(2)=9)

(c)

Brief Explanations:

We know from part (b) that (f(1)=-1) and (f(2)=9). The function (f(x)=x^{3}+3x - 5) is a polynomial (since it is of the form (a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x + a_{0}), here (n = 3), (a_{3}=1), (a_{2}=0), (a_{1}=3), (a_{0}=-5)). Since (f(1)) and (f(2)) have opposite signs ((f(1)<0) and (f(2)>0)), by the Intermediate Value Theorem for Polynomials, there is at least one (c\in(1,2)) such that (f(c)=0). Option B is wrong because (f(x)) is a polynomial. Option C is wrong because (f(1)) and (f(2)) have opposite signs.

Answer:

A. (f(x)) is a polynomial and (f(1)) and (f(2)) have opposite signs so (f(x)) has at least one (c) in the interval ((1,2)) where (f(c)=0)