Problem

Given that the Maclaurin series of $\sin x$ is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$.

(a) Explain why this fact leads to $\frac{\sin x}{x} \approx 1 - \frac{x^2}{6} + \frac{x^4}{120}$ for small positive $x$-values.

(b) Sketch the graph of $f(x) = 1 - \frac{x^2}{6} + \frac{x^4}{120}$. Label the local extrema and inflection points.