Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

Let $h (x) = f (x) - \frac{x + 1}{2} = g (x) + \sin (x) - \frac{x + 1}{2}$ . Note that since $g$ is differentiable (and since $\sin (x)$ is differentiable), then $h$ is also differentiable (and therefore continuous).

Let $c_{1} = \frac{π}{2}$ , then $h (c_{1}) = g (c_{1}) + \sin (c_{1}) - \frac{c_{1} + 1}{2} \geq \frac{c_{1}}{2} + \sin (c_{1}) - \frac{c_{1} + 1}{2} = \sin (c_{1}) - \frac{1}{2} = \frac{1}{2} > 0$ .

Let $c_{2} = - \frac{π}{2}$ , then $h (c_{2}) = g (c_{2}) + \sin (c_{2}) - \frac{c_{2} + 1}{2} \leq \frac{c_{2}}{2} + 1 + \sin (c_{2}) - \frac{c_{2} + 1}{2} = \sin (c_{2}) + \frac{1}{2} = - \frac{1}{2} < 0$ .

Thus, by the intermediate value theorem (which applies because, as we saw, $h$ is continuous), there exists $c \in (- \frac{π}{2}, \frac{π}{2})$ such that $h (c) = 0$ .

Using the same method, by the periodicity of $\sin (x)$ we see that for any integer $n$ , there exists $c \in (- \frac{π}{2} + 2 n π, \frac{π}{2} + 2 n π)$ such that $h (c) = 0$ . $So there are infinitely many real numbers c such that f (c) = \frac{c + 1}{2}$ .