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I'm trying to prove that $f(x) = x^2(\sin x)$ returns the value $y$ infinity times. I tried to use Intermediate value theorem but no success. Thanks!

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    $\begingroup$ What do you mean by "returns the value $y$ infinity times"? Do you mean it's existent everywhere? $\endgroup$ – Andrew Li Feb 26 '18 at 14:08
  • $\begingroup$ All you need is to show that $f(x)=y$ has a solution for $x>N$, for any specified $N$. To do that, show that $f$ attains arbitrarily large positive and negative values for large $x$. $\endgroup$ – lulu Feb 26 '18 at 14:10
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Let $y \in \mathbb{R}$. There are infinitely many points of the form $2n\pi + \pi/2$ for which $f(x) > y$, since at these points $f(x)$ is merely $x^2$. Similarly, we can find infinitely many points of the form $(2n+1)\pi + \pi/2$, and at these points, $f(x) = -x^2$ where $f(x) < y$.

Now the claim follows from the intermediate value theorem

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For each $x\gt \sqrt {|y|}$ you have $x^2 > |y|$.

Then for all natural $n \gt \sqrt {|y|}$ you have an interval $[2n\pi, (2n+1)\pi]$ in which $f(x) = 0$ at its endpoints and $f(x) > |y|$ at its midpoint.
Similary for interval $[(2n+1)\pi, 2(n+1)\pi]$ you have $f(x)=0$ at endpoints and $f(x)<-|y|$ at the midpoint.

Consequently, based on IVT, starting from $n$ big enough, there are at least two such $x$es in each consecutive period of sine which satisfy either $f(x)=|y|$ or $f(x)=-|y|$.

As a result there is a countably infinite set of $x$ values, which satisfy $f(x)=y$.

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Let us consider the interval $[0, \infty )$ for our domain.

$$f(x)= x^2 sin(x) $$ is the product of two functions.

1) A wave function $sin (x)$ which oscillates between $-1$ and $1$ infinitely many times.

2) An ever increasing function $x^2$ which grows up to infinity.

Thus the product is a wave function which oscillates and grows in magnitude, hitting the $x$ axis infinitely many times.

Therefore each $y$ value will be attained infinitely many times as $x$ approaches infinity.

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COMMENT.-The continuous function$f(x)=x^2\sin(x)$ goes from $-\infty$ to $+\infty$ (because for all $N\gt0$ there is $x_0\in\mathbb R$ such that $x_0^2\sin(x_0)\gt N$ which is clear; similarly with $N\lt0$ ).

Then, since $\sin(x)$ goes from $-1$ to $1$ infinitely many times do you have that the equation $$x^2\sin(x)=a\in\mathbb R$$ has infinitely many solutions.

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