Question 7

Algebra Level 4

let f : R → R f:R\rightarrow R be a bijective function and f ′ ( x ) = x 100 + 1 x 100 + 2 f'(x)=\dfrac {x^{100}+1}{x^{100}+2} . If f ( 1 ) = α f(1)=\alpha , then ( f − 1 ) ′ ( α ) = ? (f^{-1})' (\alpha) =?

∙ \bullet This question is part of the set For the JEE-nius;P

0 3 2 \dfrac {3}{2} 2 3 \dfrac {2}{3} 1

Samarpit Swain by Samarpit Swain , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

we need to find ( f − 1 ) ′ ( f ( 1 ) ) (f^{-1})'(f(1)) which can be obtained by:

f − 1 ( f ( x ) ) = x f^{-1} (f(x))=x

Using chain rule,

[ ( f − 1 ) ′ ( f ( x ) ) ] f ′ ( x ) = 1 [(f^{-1})'(f(x))]f'(x)=1

So our answer is 1 f ′ ( 1 ) = 3 2 \frac{1}{f'(1)}=\frac{3}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

It should be d/dx(inverse(f(alpha)), multiplication sign is wrong.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

everything is all right...

Kislay Raj - 6 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...