question given the polynomial f(x) = x² - 3x, what is the smallest positive integer a such that the…

question given the polynomial f(x) = x² - 3x, what is the smallest positive integer a such that the intermediate value theorem guarantees a zero exists between 1 and a? enter an integer as your answer. for example, if you found a = 8, you would enter 8. provide your answer below:

question given the polynomial f(x) = x² - 3x, what is the smallest positive integer a such that the intermediate value theorem guarantees a zero exists between 1 and a? enter an integer as your answer. for example, if you found a = 8, you would enter 8. provide your answer below:

Answer

Explanation:

Step1: Evaluate f(1)

$f(1)=1^{2}-3\times1=1 - 3=-2$

Step2: Evaluate f(a) for different positive - integer values of a

Let's start checking positive - integer values of a. For a = 2, $f(2)=2^{2}-3\times2=4 - 6=-2$ For a = 3, $f(3)=3^{2}-3\times3=9 - 9 = 0$. But we want to use the Intermediate Value Theorem, which requires $f(1)$ and $f(a)$ to have opposite signs. For a = 4, $f(4)=4^{2}-3\times4=16 - 12 = 4$ Since $f(1)=-2$ and $f(4)=4$, and the function $y = f(x)$ is a polynomial (continuous everywhere), by the Intermediate Value Theorem, there is a zero between 1 and 4.

Answer:

4