I have a function that is real valued in the interval $x\in \{-\infty,dw]$
$$ f(x)=\sqrt{-(x+w)}\sin\Bigl(b\sqrt{-(x+w)}\Bigr)\cos\Bigl(a\sqrt{-(x+dw)}\Bigr) +c\Biggl[ \sqrt{-(x+dw)}\sin\Bigl(a\sqrt{-(x+dw)}\Bigr)\cos\Bigl(b\sqrt{-(x+w)}\Bigr) \Biggr] $$
where $a$,$b$,$c$,$d$, and $w$ are all real positive constants.
If I divide $f(x)$ by $\cos\Bigl(b\sqrt{-(x+w)}\Bigr)$, I get another related function
$$ u(x)=\cos\Bigl(a\sqrt{-(x+dw)}\Bigr) \tan\Bigl(b\sqrt{-(x+w)}\Bigr) +c\Biggl[\frac{\sqrt{-(x+dw)}}{\sqrt{-(x+w)}}\sin\Bigl(a\sqrt{-(x+dw)}\Bigr) \Biggr] $$
Both of these functions share the same zeroes, and it appears (see the figure below) that the asymptotes of $u(x)$ coincide with the extrema of $f(x)$, although I would like to be able to prove this algebraically, something I'm having difficulty doing.
From the expression for $u(x)$, one can deduce that its asymptotes correspond to
$$ b\sqrt{-(x+w)}=(2n+1)\frac{\pi}{2} $$
so that the $n^{\text{th}}$ asymptote is at
$$ \bar{x}_n = -\Bigl[w+\Bigl(\frac{(2n+1)\pi}{2b}\Bigr)^2\Bigr] \forall n: \bar{x}_n < -dw$$
I got an expression for the derivative of $f(x)$ and set it to zero, but it yielded an unpromising jumble of terms that didn't help.
I'd appreciate suggestions for a clever way to go about proving this.
If it's provable, then that's extremely useful because I can then use the mid-point between asymptotes to get a very good initial guess for the zeroes of $f(x)$ which is my prime objective.
Thanks