Show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one-to-one. How would I show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one to one?
Showing that the function is strictly increasing seemed to be the way to go, but then I need to show that $f'(x) = 3x^2 + 3\cos x - 2\sin x > 0$. I graphed the derivative function and it is indeed strictly positive. Thoughts?
 A: To show that $f(x) = x^3 + 3 \sin x + 2 \cos x $ is one-to-one, it is enough to show that $ f'(x) = 3 x^2 + 3 \cos x - 2 \sin x > 0 $ for all $ x $. Notice that $ (3 \cos x - 2 \sin x) \ge -2 $ for all $ x $. So, whenever $ 3 x^2 > 2 $, $ f'(x) $ must be positive. Note that for $ x \le -\pi/2 $ or $ x \ge \pi/2 $, $ 3 x^2 \ge 3 (\pi/2)^2 > 3 (3/2)^2 > 2 $. Hence, $ f'(x) > 0 $ for $ x \le -\pi/2 $ or $ x \ge \pi/2 $. Next, for $ -\pi/2 < x < 0 $, all three terms in the expression of $f'(x)$ are positive, and so $f'(x)$ is also positive.
Finally, consider the case $ 0 \le x \le \pi/2 $. Two ways of establishing $f'(x) > 0$ is described below: one requires a calculator, the other does not.
Using calculator: Note that within $ 0 \le x \le \pi/2 $, $(3 \cos x - 2 \sin x) \le 0$ for $ \tan^{-1} (3/2) \le x \le \pi/2 $. Using a calculator (or your favorite application or programming language), verify that $3 (\tan^{-1} (3/2))^2 > 2$. Since $3 x^2 \ge 3 (\tan^{-1} (3/2))^2 > 2 $ for $ \tan^{-1} (3/2) \le x \le \pi/2 $, we have $ f'(x) > 0 $ for $ 0 \le x \le \pi/2 $.
Without using calculator: This method would not require any calculator, but only knowledge of the values of $\sin x$ and $\cos x$ for $x = (\pi/4), (\pi/3)$, which follows from basic geometry. Like before, it is sufficient to show that $ f'(x) > 0 $ for $ \tan^{-1} (3/2) \le x \le \pi/2 $. Since $\tan^{-1} x$ is a strictly increasing function in $ 0 \le x \le \pi/2 $, it follows that $ \pi/4 = \tan^{-1} 1 < \tan^{-1} (3/2) < \tan^{-1} \sqrt{3} = \pi/3 $. Note that $ 3 x^2 \ge 3 (\pi/3)^2 > 3 $ for $\pi/3 \le x \le \pi/2$, and hence $f'(x) > 0$ for $\pi/3 \le x \le \pi/2$. So, it is sufficient to show that $f'(x) > 0$ for $\pi/4 \le x \le \pi/3$ to complete the proof. Note that for $\pi/4 \le x \le \pi/3$, $3x^2$ is an increasing function, while $(3 \cos x - 2 \sin x)$ is a decreasing function. Therefore, for $\pi/4 \le x \le \pi/3$, $f'(x) \ge$ $ 3 (\pi/4)^2 + (3 \cos (\pi/3) - 2 \sin (\pi/3))$ $> 3 (3/4)^2 + ((3/2) - \sqrt{3}) = (51/16) - \sqrt{3} > 3 - \sqrt{3} > 0$.
