I am struggling with the following exercise:
Do the two equations $$x+y-\sin(z) = 0 \text{ and }e^x-x-y^3=1$$ define two functions $y(x), z(x)$ in a neighbourhood of $x=0$ such that $y(0)=z(0)=0$. If so examine if these functions have local extrema at $x=0$.
My attempt: My idea would be to use that $y(x) = \sqrt[3]{e^x-x-1}$, which implies $z(x) = \sin^{-1}(x+\sqrt[3]{e^x-x-1})$ . But these functions are very complicated. So I think that the Implicit Function Theorem would be better suited for this case. I have read about it on wikipedia, but I am not quite sure how to use it yet. Could you help me?
Define $F:\mathbb R^3\to \mathbb R^2$ by $F(x,y,z)=(x+y-\sin(z), e^x-x-y^3-1)$. Then, since
$\mathcal JF(y,z)=\begin{pmatrix} 1 & -\cos z \\ 3y^2& 0 \end{pmatrix}$ so that $\det \mathcal JF(0,0)=0,$ the implicit function theorem does not give any information.
On the other hand, you have directly,
$y'(x)=\dfrac{\mathrm{e}^x-1}{3\left(\mathrm{e}^x-x-1\right)^\frac{2}{3}}$ so it is easy to see that $y$ increases on $(0,\infty)$ and decreases on $(-\infty,0)$, so has a global minimum there because $y>0$ on $\mathbb R$.
Note, $y$ is not differentiable at $x=0$, which is why the implicit function theorem did not give an affirmative answer.
Now, $z'(x)=\dfrac{\frac{\mathrm{e}^x-1}{3\left(\mathrm{e}^x-x-1\right)^\frac{2}{3}}+1}{\sqrt{1-\left(\sqrt[3]{\mathrm{e}^x-x-1}+x\right)^2}}$, which looks complicated, but it's not hard to see that in a sufficiently small neighborhood $(-\delta,\delta)$ of $x=0,\ z'((-\delta,0))<0$ and $z'((0,\delta))>0$ so $z$ has a local minimum at $x=0.$ Note, as in the case for $y$, that $z'(0)$ does not exist.