Final answer:
To demonstrate that the function g(x) = x^2 - x - sin(x) has a root between 1 and 2, we apply the Intermediate Value Theorem after calculating that g(1) is negative and g(2) is positive, which means there exists at least one value c in (1,2) such that g(c) = 0.
Step-by-step explanation:
To show that the function g(x) = x2 - x - sin(x) has a root between 1 and 2, we can use the Intermediate Value Theorem. This theorem states that if a function f is continuous on a closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one number c in the interval (a, b) such that f(c) = N.
First, we evaluate the function at g(1) and g(2). If g(1) is less than zero and g(2) is greater than zero, it implies by the Intermediate Value Theorem that there must be some value c in the interval (1,2) where g(c) = 0. Let's compute:
- g(1) = 12 - 1 - sin(1) = 1 - 1 - sin(1) = -sin(1)
- g(2) = 22 - 2 - sin(2) = 4 - 2 - sin(2) = 2 - sin(2)
Since sin(x) is always less than or equal to 1, g(1) is negative and g(2) is positive. Thus, there is at least one root in the interval (1,2).