Assuming the problem as stated is, given
[tex]f(x)=\displaystyle\int_3^x\sqrt{1+t^3}\,\mathrm dt[/tex]
find [tex](f^{-1})'(0)[/tex].
Provided that [tex]f[/tex] is (at least locally) invertible, we have [tex]f(f^{-1}(x))=x[/tex], and differentiating both sides with respect to [tex]x[/tex] gives
[tex]f'(f^{-1}(x))(f^{-1})'(x)=1\implies (f^{-1})'(x)=\dfrac1{f'(f^{-1}(x))}[/tex]
Notice that [tex]f(3)=0[/tex], which means [tex]f^{-1}(0)=3[/tex]. From this it follows that
[tex](f^{-1})'(0)=\dfrac1{f'(f^{-1}(0))}=\dfrac1{f'(3)}[/tex]
and since [tex]f'(x)=\sqrt{1+x^3}[/tex] by the fundamental theorem of calculus, it follows that [tex]f'(3)=2\sqrt7[/tex], and so [tex](f^{-1})'(0)=\dfrac1{2\sqrt7}[/tex].
