Integral (3,x) of sqrt ( 1+t^3) find (f^-1)'(0)

Question

asked Feb 23, 2018 205k views

1 Answer

Stroz · Answer 1 · 2018-02-26T17:41:38+0000

Assuming the problem as stated is, given

$f(x)=\displaystyle\int_3^x√(1+t^3)\,\mathrm dt$

find
(f^(-1))'(0)

.

Provided that

is (at least locally) invertible, we have
f(f^(-1)(x))=x

, and differentiating both sides with respect to

gives

$f'(f^(-1)(x))(f^(-1))'(x)=1\implies (f^(-1))'(x)=\frac1{f'(f^(-1)(x))}$

Notice that
f(3)=0

, which means
f^(-1)(0)=3

. From this it follows that

$(f^(-1))'(0)=\frac1{f'(f^(-1)(0))}=\frac1{f'(3)}$

and since
f'(x)=√(1+x^3)

by the fundamental theorem of calculus, it follows that
$f'(3)=2\sqrt7$ , and so
$(f^(-1))'(0)=\frac1{2\sqrt7}$ .