Question

Use the Extreme Value Theorem to prove that f(x) = x² - x + 1 + cos(x) has a minimum value on R.

It might be helpful to start by completing the square.
You won't be able to immediately determine the exact value of the minimum, but you can prove that it exists.

Answer 1

The Extreme Value Theorem states that if a function is continuous on a closed interval, it must have a minimum and a maximum value. By completing the square, we can find the minimum value of f(x) = x² - x + 1 + cos(x) on the entire real line R to be 0.75 when x = 0.5.

Step-by-step explanation:

The Extreme Value Theorem states that if a function f(x) is continuous on a closed interval [a, b], then it must have a minimum and a maximum value on that interval. To prove that f(x) = x² - x + 1 + cos(x) has a minimum value on the entire real line R, we can show that the function is continuous and unbounded below.

First, let's complete the square for the quadratic term x² - x + 1 to get f(x) = (x - 0.5)² + 0.75 + cos(x). Since the cosine function is bounded between -1 and 1, the minimum value of f(x) will occur when the quadratic term is at its minimum value.

The quadratic term is a perfect square, and its minimum value is 0 when x = 0.5. Therefore, the minimum value of f(x) is 0.75 when x = 0.5.