Question

The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval.. . sin(x)= x^2-x, (1,2). . Can someone see if this is correct?. . f(x) = sin(x) - x^2 + x, then f is continuous since sin(x) and x^2 - x are continuous and their composition is also continuous. sin(x) = x^2 - x which is equivalent to showing that f(x) = 0. f(1) = 0.841 and f(2) = -1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0. there is a root to the equation. sin(x) = x^2 - x.

Answer 1

Base on the equation given above the intermediate value theorem shows the correct answer, all polynomials are continuous. Sine is continuous making it sin(x) - x ^2 + x is continuous so f(1) = 0.841 and f(2) = -1.091. The sign changes making f(x) continuous.