Well, if #g(x) = f^(-1)(x)#, then #f(g(x))=x# by the definition of an inverse function.
#f(g(x)) = g^3(x) +e^(g(x)/2) = x#
#g^3(x) = x-e^(g(x)/2)#
In order to find #g'(x)#, let's take the derivative of both sides.
#[g^3(x)]' = [x-e^(g(x)/2)]'#
For the first one, we have
#g^3(x) = g(x)g(x)g(x)#
We have to use the #color(blue)("Product rule")# twice:
#:. [g^3(x)]'=[g(x)g(x)]'g(x) + g(x)g(x)g'(x)#
#[g(x)g(x)]' = g'(x)g(x)+g(x)g'(x) = 2g'(x)g(x)#
#=> [g^3(x)]' = 2g^2(x)g'(x) + g^2(x)g'(x) = 3g^2(x)g'(x)#
For the second part:
#[x-e^(g(x)/2)]' = [x]' - [e^(g(x)/2)]'= 1 - [e^(g(x)/2)]'#
Now, let #alpha = e^(g(x)/2)#.
#alpha = e^(g(x)/2)#
Take the natural logarithm of both sides.
#ln alpha = g(x)/2#
Differentiate both sides.
#(alpha')/alpha = (g'(x))/2=> alpha' = (g'(x))/2alpha#
Let's undo the substitution:
#[e^(g(x)/2)]' = (g'(x))/2e^(g(x)/2)#
Thus,
#[x-e^(g(x)/2)]' = 1-(g'(x))/2e^(g(x)/2)#
#color(blue)( :. 3g^2(x)g'(x) = 1-(g'(x))/2e^(g(x)/2)#
Plug in #x=1# and get:
#3g^2(1)color(Red)(g'(1))=1-color(red)(g'(1))/2e^(g(1)/2)#
Now, we have to find #g(1)#. Let's use our original identity, that
#f(g(x))=x#, for all #x#.
#f(g(1))=1#
#g^3(1)+e^(g(1)/2) = 1#
To make this equality simpler to look at, let #u = g(1)#.
#u^3 + e^(u/2) = 1#
Well, you might be able to see that #u=0# is a solution of this equation. But how do we prove it's the only one?
Well, it's clear that the function is monotonically increasing. This is because, as #x# gets bigger and bigger, so does #f(x)#. This makes #u=0# the only solution.
#:. color(blue)(g(1)=0)#.
Finally, we have:
#3*0*color(Red)(g'(1))=1-color(red)(g'(1))/2e^0#
#1-color(red)(g'(1))/2 = 0#
#=> color(Red)(g'(1)=2)#.
