is a monotonic function multiplied by an invertible function still invertible?

Question

If we have an invertible function $f(x)$ defined on the Reals, and we have a strictly monotonic function $m(x)$ also defined on the Reals.

Is $g(x):=m(x)f(x)$ then also an invertible function? Is there a simple proof for this?

Edit: What if we additionally assume $m(x)$ is strictly positive, like $m(x)=e^x$

Answer 1

Not necessarily, since the rates at which $m(x)$ and $f(x)$ change can vary. For example, suppose $m(x)$ and $f(x)$ have zeroes at two different points. Then $m(x)f(x)$ has zeroes at both of those points, and since $m(x)f(x)=0$ for two different values of $x$, the function is not one-to-one and does not have an inverse.