Question

How can I prove that the function `f(x) = x^2*(sinx)` returns the value `y` infinity times?

Answer 1

Here are some thoughts that should help. For a fully detailed proof, please see the answer from Andrea S.

Explanation:

There are probably several good ways to prove this.

Note that as `x` varies from an even multiple of `pi` to an odd multiple of `pi`, the values of `y=f(x)` vary from `0` to `x^2` and back to `0`.

So, the case `y = 0` is done.

For positive `y`, there is an `X` such that, for `x >= X`, we have `x^2 > y`.

Once we have reached this `X`, there are infinitely many more intervals of the form `[2kpi, (2k+1)pi]` and on each such interval `y` is twice.
On each such interval, use the Intermediate Value Theorem to conclude that there are two `x` with `f(x) = y`
Then, since there are infinitely many such intervals, there must be infinitely many such `x`'s.

For `y < 0` modify that argument above using intervals of the form `["odd"xx pi, "even"xx pi]`

Answer 2

Note first that for `y=0` the proposition is true as:

`f(kpi) = (kpi)^2sin(kpi) = 0`

for any `k in ZZ`.

Now, given any value `y in RR` with `y > 0` consider the sequence: `{a_n = 2kpi+pi/2}` and the intervals:

`I_k = [2kpi, 2kpi+pi/2]`

with `k in NN^0`

As `lim_(k->oo) a_k= +oo` we must be able to find an integer `N` such that:

`k >=N => a_k > sqrty`

Now as `f(x)` is continuous in the interval, `I_N` and:

`f(2Npi) = (2Npi)^2 sin(2Npi) = 0`

`f(2Npi+pi/2) = (2Npi+pi/2)^2 sin(2Npi+pi/2) = a_N^2`

then as `x` varies in the interval `I_N`, `f(x)` must assume all the values in the interval `(0,a_n^2)`, including `y` because by means of how we chose `N` we have:

`0 < y < a_N^2`

But similarly for `x in I_(N+1)`, `f(x)` must assume all the values between `0` and `a_(N+1)^2 > a_N^2` which include again `y`.

We can therefore conclude that for every `k>=N` ther must be a point `xi_k in I_k` such that `f(xi_k) = y` which proves the point.

If we choose `y < 0` then we can apply the same demonstration choosing as intervals `J_k = [2kpi-pi/2, 2k]`.