Skip to Content

Intermediate Value Theorem Root Proof

Home | Real Analysis | Limits and Continuity of Functions | Intermediate Value Theorem Root Proof

Use the Intermediate Value Theorem to show that there is a root of the equation f(x)=x2x12f(x) = x^2 - x - 12 in the interval [3, 5].

The Intermediate Value Theorem is an essential concept in real analysis, particularly related to understanding the behavior of continuous functions over an interval. At its core, the theorem asserts that if a function is continuous on a closed interval and takes different signs at the endpoints of that interval, then it must cross zero somewhere within that interval. This concept leverages continuity to guarantee the existence of a root, providing a bridge between algebraic function behavior and geometric intuition.

In this problem, we are asked to demonstrate that the polynomial function f(x)=x2x12f(x) = x^2 - x - 12 has at least one real root within the given interval from 3 to 5. Since polynomial functions are continuous over real numbers, you can confidently apply the Intermediate Value Theorem. Begin by evaluating the function at the endpoints of 3 and 5. If the function takes on values with opposite signs at these points, the theorem guarantees a root exists between them.

This approach not only reinforces the application of the Intermediate Value Theorem but also strengthens your understanding of how continuity can be a powerful tool in establishing the existence of solutions to equations. Furthermore, this problem invites exploration into how such theorems lay the foundation for more advanced topics in calculus and analysis, such as derivative tests and advanced function behaviors.

Posted by Gregory 4 hours ago

Related Problems

Prove that limx3(2x+1)=5\lim_{{x \to 3}} (-2x + 1) = -5 using an epsilon-delta proof.

Prove that the function f(x)=xf(x) = |x| is continuous on the real numbers using the epsilon-delta definition.

Use the Intermediate Value Theorem to find the value of cc in the interval [1,4][1, 4] such that f(c)=19f(c) = 19, given f(x)=2x2+3x+5f(x) = 2x^2 + 3x + 5.