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If $f: \mathbb R \to \mathbb R$ defined by $f(x)= x^3+ px^2 + qx+ k \sin{x}$, where $k,p,q \in \mathbb R$.

Find the condition for $f^{-1}$ to exist.

Can somebody please give me a Hint how to solve this problem.

EDIT(After getting hints): If we can show that $f$ is strictly increasing, then we can get that $f$ is injective. So we want $f^{'}(x)>0$ and that gives me $p^2 < 3(q + k \cos{x}).$

As $f^{'}(x)$ will be a quadratic polynomial and we want it to be strictly positive or strictly negative but as coefficient of $x^2$ is positive, we can only get derivative to be strictly positive if discriminant is $<0.$

User
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  • $f$ needs to be a bijection. – ervx Jul 14 '17 at 20:21
  • @ervx, that much I know. How to find conditions for that is a question. – User Jul 14 '17 at 20:24
  • @MercyKing That's not really a hint, is it? – Jason Jul 14 '17 at 20:42
  • @MercyKing. It's okay if $f'(x) = 0$ at discrete points. For example, consider $x^3$. But $f$ must be strictly monotone. – md2perpe Jul 14 '17 at 20:45
  • The condition is $f'(x)\ge 0$ or $f'(x)\le 0$ for all $x\in \mathbb{R}$, or equivalently $p^2-3q\le 0$. – HorizonsMaths Jul 14 '17 at 20:52
  • @MercyKing, don't we want the derivative to be strictly positive? – User Jul 14 '17 at 21:02
  • @User derivative to be strictly positive? No, it's enough for the derivative to be $\ge 0$ with only isolated zeros. Take for example $y = x^3$ which is strictly increasing, yet its derivative is $y'=0$ at $x=0$. – dxiv Jul 14 '17 at 23:19
  • @dxiv, but if we are taking derivative to be $\ge 0$, how can we eliminate the case of constant function or somethibg like that? – User Jul 15 '17 at 01:44
  • @User Emphasis on "only isolated zeros". Actually, it's enough that the interior of the zero set of the derivative be empty, see for example Monotone functions and non-vanishing of derivative. – dxiv Jul 15 '17 at 02:35
  • We want $\forall x, 3x^2+2px + q - k \cos x \geq 0$, since $3x^2+2px+q-k \cos x \geq 3x^2 + 3px +q -|k| $, a sufficient condition is $p^2 \leq 3(q-|k|)$. can't figure out necessary condition though. – Siong Thye Goh Jul 15 '17 at 02:59
  • @SiongThyeGoh, How did you get $3px$ in the second equation? Is that a typo? – User Jul 15 '17 at 03:13
  • Yikes, it is indeed a typo. should be $2px$. – Siong Thye Goh Jul 15 '17 at 03:18
  • @dxiv, I agree with your argument. So, you mean, to check whether a given function is strictly increasing or decreasing we need to check that $f^{'}(x)\ge 0$ or $f^{'}(x)\le 0$ and then we need to verify that $f^{'}(x)= 0$ is on isolated points only. Am I right? – User Jul 15 '17 at 03:19
  • @User That is sufficient, indeed. The weaker condition derived in the linked post is enough, though, which is that the set of zeros of the derivative $,{x \in \mathbb{R} \mid f'(x) = 0 },$ has an empty interior. – dxiv Jul 15 '17 at 03:48

2 Answers2

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Guide:

Check $\lim_{x \rightarrow \infty} f(x)$.

Check $\lim_{x \rightarrow -\infty} f(x)$.

Note that $f$ is continuous.

Find conditions to make it strictly increasing.

Siong Thye Goh
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Strictly increasig functions are injective, and hence, have inverse.

ajotatxe
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