Question

using the intermediate value theorem, does the polynomial ( f(x) = x^3 - 9x ) have at least one zero between ( x = -4 ) and ( x = -2 )?
yes, there is at least one zero in the interval (-4, -2).
there is not enough information to tell.

Explanation:

Step1: Recall Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f(x) \) is continuous on a 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 zeros, we check if \( f(a) \) and \( f(b) \) have opposite signs (since \( k = 0 \) will be between them if \( f(a) \) and \( f(b) \) have opposite signs). Polynomial functions are continuous everywhere, so \( f(x)=x^{3}-9x \) is continuous on \([-4, -2]\).

Step2: Calculate \( f(-4) \)

Substitute \( x=-4 \) into \( f(x) \):
\( f(-4)=(-4)^{3}-9\times(-4) \)
\( = -64 + 36 \)
\( = -28 \)

Step3: Calculate \( f(-2) \)

Substitute \( x = -2 \) into \( f(x) \):
\( f(-2)=(-2)^{3}-9\times(-2) \)
\( = -8 + 18 \)
\( = 10 \)

Step4: Analyze signs of \( f(-4) \) and \( f(-2) \)

\( f(-4)=-28 \) (negative) and \( f(-2)=10 \) (positive). Since \( 0 \) is between \( -28 \) and \( 10 \), by the Intermediate Value Theorem, there exists at least one \( c \) in \([-4, -2]\) such that \( f(c) = 0 \).

Answer:

Yes, there is at least one zero in the interval \([-4, -2]\).