Question

1. (10 points) answer each question below.

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

Explanation:

(a)

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.

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

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:

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

(b)