Behaviour of roots of a polynomial with function coefficients

Question

Let $(-1+c_4(h))x^4 +c_3(h)x^3+c_2(h)x^2+c_1(h)x+c_0(h)=0$ be an equation with variable coefficients, depending smoothly on $h$. Also let $0\le c_4(h)\le 1-\epsilon$ for some $\epsilon>0$ and $c_0(h)>\epsilon'$ for some $\epsilon'>0$. One can easily see that for any $h$, this equation has two real solution, one positive and one negative. Let $z(h)$ be the positive roots. The question is that under what conditions I can say that $z(h)$ is a differentiable function of $h$?

Answer 1

In fact, you can have one or two positive roots. Using the IFT, $$ 0\ne\frac{\partial}{\partial x}(-1+c_4(h))x^4+c_3(h)x^3+c_2(h)x^2+c_1(h)x+c_0(h)\Big\vert_{x=z(h)} $$ $$ =4(-1+c_4(h))x^3+3c_3(h)x^2+2c_2(h)x+c_1(h)x\Big\vert_{x=z(h)} $$ while $$(-1+c_4(h))x^4 +c_3(h)x^3+c_2(h)x^2+c_1(h)x+c_0(h)\vert_{x=z(h)}=0,$$ i.e.,the root(s) $z(h)$ ($z_1(h)$, $z_2(h)$) is (are) simple.