$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$
Let $$g\left( x \right) = \int\limits_0^{{e^x}} {{{f'\left( t \right)} \over {1 + {t^2}}}} \,dt.$$
Which of the following is true?
Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$
$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$
If $$f''\left( x \right) < 0\,\forall x \in \left( {a,b} \right)$$ and $$c$$ is a point such that $$a < c < b,$$ and
$$\left( {c,f\left( c \right)} \right)$$ is the point lying on the curve for which $$F(c)$$ is
maximum, then $$f'(c)$$ is equal to