If F(x) is a differentiable function, and #F(-1) = 5#, #F'(-1) = 4# and #(F^-1)(4) = 2/3# then what is #(F^-1)'(5)#?

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Answer 1 · 2017-01-22T17:57:21

#(F^-1)'(5)=1/4#.

#(F^-1@F)(x)=x rArr F^-1(F(x))=x#

Diff.ing both the sides w.r.t #x# using the Chain Rule, then,

#(F^-1)'((F(x))F'(x)=1#.

#"Taking, "x=-1," we get, "(F^-1)'(F(-1))F'(-1)=1#.

Substituting the given values, we get,

#{(F^-1)'(5)}(4)=1#

#:. (F^-1)'(5)=1/4#

Taking the other way round, #(F@F^-1)(x)=x#

#:. F(F^-1(x))=x rArr d/dx{F(F^-1(x))}=d/dx(x)#

#rArr F'(F^-1(x)){F^-1(x)}'=1#

#"Taking "x=5, F'(F^-1(5){F^-1(5)}'=1...............(star)#

Recall that #F(-1)=5 rArr F^-1(5)=-1#

#:. F'(F^-1(5))=F'(-1)=4," and, so, by "(star),# we have,

#4{F^-1(5)}'=1 :. (F^-1)'(5)=1/4#.