Use the intermediate value theorem for polynomials to show that the polynomial function has a real zero between the numbers given.
-$f ( x ) = 3 x ^ { 3 } - 3 x + 8 ; - 2$ and $- 1$
A) $f ( - 2 ) = - 10$ and $f ( - 1 ) = 8$
B) $f ( - 2 ) = - 9$ and $f ( - 1 ) = 9$
C) $f ( - 2 ) = 10$ and $f ( - 1 ) = - 8$
D) $f ( - 2 ) = 10$ and $f ( - 1 ) = 8$

Question

Quizplus · Accepted Answer

The Intermediate Value Theorem states that if $f(x)$ 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 $f(c) = N$. Calculating $f(-2) = 3(-2)^3 - 3(-2) + 8 = -24 + 6 + 8 = -10$ and $f(-1) = 3(-1)^3 - 3(-1) + 8 = -3 + 3 + 8 = 8$, we see that the function values at $-2$ and $-1$ are $-10$ and $8$, respectively. Since the function values change sign from negative to positive, there must be at least one real zero between $-2$ and $-1$.

Use the Intermediate Value Theorem for Polynomials to Show That f(x)=3x3−3x+8;−2f ( x ) = 3 x ^ { 3 } - 3 x + 8 ; - 2f(x)=3x3−3x+8;−2

Use the Intermediate Value Theorem for Polynomials to Show That $f ( x ) = 3 x ^ { 3 } - 3 x + 8 ; - 2$