Evaluating an implicit function Consider the functions defined implicitly by $y^3 - 3y + x = 0$. If $x \in (-\infty,-2) \cup (2, \infty), y=f(x).$  If $x \in (-2,2), y= g(x)$. Also, $g(0)=0$. 
Then,


*

*If $f(-10 \sqrt{2}) = 2 \sqrt{2}$, then $f''(-10\sqrt{2})$ = 

*$\int_{-1}^{1}{g'(x)}$ =
 A: Consider the function
$$p:\quad y\mapsto x:=3y-y^3\qquad(-\infty<y<\infty)\ ;$$
its graph is a cubic parabola. The function $p$ is odd,  has a local minimum at $(-1,-2)$, a local maximum at $(1,2)$, and is monotonic otherwise. There is a unique inverse $f:\>x\mapsto y=f(x)$ for $-\infty<x<-2$ and for $2<x<\infty$, whereas for $-2<x<2$ the set $p^{-1}(\{x\})$ has three elements. Now $p$ maps the interval $J:=\ ]{-1},1[\ $ bijectively onto $\ ]{-2},2[\ $, so that the restriction $p\restriction J$ has a well defined inverse $g:\>\ ]{-2},2[\ \to\ ]{-1},1[\ $.


*

*Since $p(2\sqrt{2})=-10\sqrt{2}<-2$ we have $f(-10\sqrt{2})=2\sqrt{2}$. In order to compute $f''(-10\sqrt{2})$ we start with
$$f'(x)={1\over p'\bigl(f(x)\bigr)}$$ and obtain
$$f''(x)=-{1\over p'^2\bigl(f(x)\bigr)}p''\bigl(f(x)\bigr)f'(x)=-{p''\bigl(f(x)\bigr)\over p'^3\bigl(f(x)\bigr)}\ .$$
It follows that
$$f''(-10\sqrt{2})=-{p''(2\sqrt{2})\over p'^3(2\sqrt{2})}=-{4\sqrt{2}\over 3087}\ .$$

*By the FTC one has
$$\int_{-1}^1 g'(x)=g(1)-g(-1)=2g(1)\doteq0.694593\ ,$$
whereby the last value came from numerically solving $y^3-3y+1=0$ and choosing the root in $[0,1]$.

