Use the intermediate value theorem for polynomials to show that the polynomial function has a real zero between the
numbers given.
-$$f ( x ) = 6 x ^ { 3 } - 10 x + 7 ; - 2 \text { and } - 1$$
A) $f ( - 2 ) = - 21$ and $f ( - 1 ) = 11$
B) $f ( - 2 ) = - 20$ and $f ( - 1 ) = 12$
C) $f ( - 2 ) = 21$ and $f ( - 1 ) = - 11$
D) $f ( - 2 ) = 21$ and $f ( - 1 ) = 11$

Question

Quizplus · Accepted Answer

The intermediate value theorem states that if a function $f$ is continuous on a closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $[a, b]$ such that $f(c) = N$. For the polynomial $f(x) = 6x^3 - 10x + 7$, we calculate $f(-2)$ and $f(-1)$ to find the values at the endpoints of the interval $[-2, -1]$. Calculating $f(-2) = 6(-2)^3 - 10(-2) + 7 = -48 + 20 + 7 = -21$ and $f(-1) = 6(-1)^3 - 10(-1) + 7 = -6 + 10 + 7 = 11$. Since $f(-2)$ and $f(-1)$ have opposite signs, there must be at least one real zero between $-2$ and $-1$ according to the intermediate value theorem.

Use the Intermediate Value Theorem for Polynomials to Show That f(x)=6x3−10x+7;−2 and −1f ( x ) = 6 x ^ { 3 } - 10 x + 7 ; - 2 \text { and } - 1f(x)=6x3−10x+7;−2 and −1

Use the Intermediate Value Theorem for Polynomials to Show That $f ( x ) = 6 x ^ { 3 } - 10 x + 7 ; - 2 \text { and } - 1$