To determine whether the function [tex]\( f(x) = x^3 + 5x^2 - 9x - 37 \)[/tex] has at least one real zero between [tex]\( a = -9 \)[/tex] and [tex]\( b = -3 \)[/tex] using the Intermediate Value Theorem, we proceed as follows:
1. Evaluate the Function at the Endpoints:
Calculate [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex].
- For [tex]\( a = -9 \)[/tex]:
[tex]\[
f(-9) = (-9)^3 + 5(-9)^2 - 9(-9) - 37
\][/tex]
Simplifying step-by-step:
[tex]\[
(-9)^3 = -729
\][/tex]
[tex]\[
5(-9)^2 = 5 \times 81 = 405
\][/tex]
[tex]\[
-9(-9) = 81
\][/tex]
Putting it all together:
[tex]\[
-729 + 405 + 81 - 37 = -280
\][/tex]
Thus, [tex]\( f(-9) = -280 \)[/tex].
- For [tex]\( b = -3 \)[/tex]:
[tex]\[
f(-3) = (-3)^3 + 5(-3)^2 - 9(-3) - 37
\][/tex]
Simplifying step-by-step:
[tex]\[
(-3)^3 = -27
\][/tex]
[tex]\[
5(-3)^2 = 5 \times 9 = 45
\][/tex]
[tex]\[
-9(-3) = 27
\][/tex]
Putting it all together:
[tex]\[
-27 + 45 + 27 - 37 = 8
\][/tex]
Thus, [tex]\( f(-3) = 8 \)[/tex].
2. Apply the Intermediate Value Theorem (IVT):
The Intermediate Value Theorem states that if [tex]\( f \)[/tex] is continuous on the interval [tex]\([a, b]\)[/tex] and [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] have opposite signs, then there exists at least one [tex]\( c \)[/tex] in the interval [tex]\((a, b)\)[/tex] such that [tex]\( f(c) = 0 \)[/tex].
In this case, [tex]\( f(-9) = -280 \)[/tex] and [tex]\( f(-3) = 8 \)[/tex]. Since [tex]\(-280\)[/tex] and [tex]\(8\)[/tex] have opposite signs, there is a change in sign from negative to positive. Therefore, by the Intermediate Value Theorem, there must be at least one root (zero) between [tex]\( a = -9 \)[/tex] and [tex]\( b = -3 \)[/tex].
Thus, the correct choice is:
[tex]\[
\text{A. By the intermediate value theorem, the function has at least one real zero between } a \text{ and } b \text{ because } f(a)= -280.00 \text{ and } f(b)= 8.00
\][/tex]
