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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?

amWhy
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TheSenate
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    Since $f'(x) > 0$ for $0 \le x \le \pi/2$, and $3 (\pi/2)^2 > 7$, you have $f'(x) > 0$ for $x>0$. Similarly, notice that $f'(x) > 0$ for $-\pi/2 \le x \le 0$, and hence it is always positive. – joy Jan 19 '22 at 20:25
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    With respect to the comment of @joy, which I agree with, what is the approximate range of $x$, within $-\pi/2 \leq x \leq \pi/2$ such that $3\cos(x) \geq 2\sin(x)$? Beyond this range, with $f'(x)$ containing the added term of $3x^2$, is it possible for $f'(x)$ to be $\leq 0$? Further, if $x < -\pi/2,$ or $x > \pi/2$, considering that the cosine and sine functions are bounded, and considering that $f'(x)$ contains the $3x^2$ term, is it then possible for $f'(x)$ to be $\leq 0$? – user2661923 Jan 19 '22 at 20:35
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    Beyond $-\pi/2$ and $\pi/2$, as $3x^2$ keeps on increasing, its minimum value is $3(\pi/2)^2 > 27/4$. So, $f'(x)>0$ for $x \le -\pi/2$ and $x \ge \pi/2$. For $-\pi/2 < x < 0$, all the terms are positive, and hence their sum $f'(x)$ is positive. Now, for $0<x<\pi/2$, $3\cos x - 2\sin x \le 0$ for $x \ge \arctan(3/2)$. It can be easily checked using a calculator that $3(\arctan(3/2))^2 > 2$, and then the earlier argument regarding $3x^2$ would suffice to show that $f'(x)>0$ for $0<x<\pi/2$. – joy Jan 19 '22 at 21:10
  • Are there any reasons why this particular function is interesting? –  Jan 19 '22 at 21:18
  • @banana I suspect that the point of the problem is that it is easy for a new-intermediate student to go off the rails. As previous comments indicate, it is not that difficult to prove that $f'(x)$ is always $> 0$. However, if you shy away from that approach, it is easy to chase your tail, (unsuccessfully) trying to evaluate the function for various values of $x$. – user2661923 Jan 20 '22 at 00:35
  • @joy: I don't see why $f'(x)>0$ for $0\le x\le \pi/2$ is obvious. –  Jan 20 '22 at 00:54
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    Alright, I am posting a detailed answer. I refrained from answering the question due to this site's policy of discouraging users from answering what the moderators consider "problem statement questions" (https://math.meta.stackexchange.com/q/33508/947444). Personally, I find this policy very confusing and totally subjective. Any well-posed question, which may be considered as a "problem statement question" by one individual, can be challenging to another. – joy Jan 20 '22 at 19:06
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    As several users have discussed this question and its possible solution in the comments, I hope that it will not be considered a "problem statement question" by a moderator. – joy Jan 20 '22 at 19:06
  • "I don't see why f′(x)>0 for 0≤x≤π/2 is obvious. –" Well in the first quadrant $3\cos x = 2\sin x$ when $9\cos^2 x =4\sin^2 x$ when $13\cos^2 x=4(\sin^2x +\cos^2x)=4$ when $x = \arccos \frac {2{\sqrt 13}}\approx 0.98$. so to have $0\le x\le \frac \pi 2$ and $3\cos x<2\sin x$ we must have $0.98< x \le \frac \pi 2$ in which case $3x^2>0.98^2\times 3\approx 2.88$. As $3\cos x -2\sin x > -2\sin x > 2$ we must have $3x^2 + 3\cos x - 2\sin x > 0.88 >0$ for all $x: 0.98 < x \le \frac \pi 2$. And for $0\le x < 0.98$ we have $3\cos x > 2\sin x$ so $3x^2+3\cos x - 2\sin x > 3x^2\ge 0$. – fleablood Jan 20 '22 at 20:10
  • You can find a parabolic lower bound to $3\cos x-2\sin x$ with its apex at the first positive minimum, $x=\arctan\frac23$ (take $\alpha$ such that $\cos t<1-\alpha t^2$). Adding $3x^2$, you get a strictly positive parabola. In the other intervals where the sinusoid is negative, you have $3x^2>\sqrt{13}$ and you are safe. –  Jan 20 '22 at 20:22

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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))$ $> 3 (3/4)^2 + ((3/2) - \sqrt{3}) = (51/16) - \sqrt{3} > 3 - \sqrt{3} > 0$.

joy
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    Of course the proof can be significantly shortened. However, I tried to post an answer as detailed and as straight-forward as I could. – joy Jan 20 '22 at 19:10
  • Hi Joy. Thank you for your answer. Your original comment was more than enough for me to write out a detailed solution. I merely needed a direction. I absolutely agree with your sentiments about "problem statement questions." I simply had a math question, and you provided a hint which helped me solve it. That is what this site should be about, and I appreciate users like yourself who seek to make the site a better place. – TheSenate Jan 21 '22 at 01:15
  • Thank you @TheSenate for your words, and you are very much welcome! – joy Jan 21 '22 at 06:59