Question

why can the intermediate value theorem be used to show that the equation has a solution on (-1,5)?
a. it can be used because f(x)=x^3 - 6x^2 + 2x is defined on (-1,5) and -2 < f(-1) < f(5)
b. it can be used because f(x)=x^3 - 6x^2 + 2x is continuous on -1,5 and the function is defined at x = -1 and x = 5
c. it can be used because f(x)=x^3 - 6x^2 + 2x is defined on (-1,5) and f(-1) < -2 < f(5)
d. it can be used because f(x)=x^3 - 6x^2 + 2x is continuous on -1,5 and -2 is between f(-1) and f(5)

Explanation:

The Intermediate Value Theorem states that if a function \(y = f(x)\) 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 the open interval \((a,b)\) such that \(f(c)=k\). For the function \(f(x)=x^{3}-5x^{2}+2x\), to use the Intermediate Value Theorem to show that the equation has a solution on \((- 1,5)\), the function must be continuous on \([-1,5]\) and the value we are looking for must be between \(f(-1)\) and \(f(5)\).

Answer:

A. It can be used because \(f(x)=x^{3}-5x^{2}+2x\) is continuous on \([-1,5]\) and \(-2\) is between \(f(-1)\) and \(f(5)\)