# Given f=e^x+x+15 what is (f^-1)'(1).

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f= e^x + x + 1

First let us find the derivative:

f'(x) = e^x + 1

Now we know that:

f'(x) * (f^-1)' = -1

Now substitute:

(e^x+1) *(f^-1)' = -1

==> f^-1)' = -1/(e^x + 1)

Now to find (f^-1)'(1) we will substitute with x= -1:

==> (f^-1)(1) = -1/(e^1 + 1)

= -1/(e+1)

==> (f^-1)(1) = -1/(e+1)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = e^x+x+15.

f(x) is a continuous monotonosly increasing function as e^x and x+15 are also continous monotonously incresing. So f(x) = e^x+x+15 is a bijection . So the f^-1 function exists.

To find  f^(-1)'(1).

f(x) = e^x+x+15,

Let y = f^(-1)x.

Then f(y) = x.

x = e^y+y+15 by definition

Differentiate with respect y.

dx/dy = e^y+1.

Take the reciprocal .

dy/dx = 1/(e^y+1)

(f^-1)' (x) = 1/e^y +1

(f^'-1) (x)= 1{ e^(e^x+x+15) +1}

(f^-1)'(-1) = 1/{e^(e+1+15) +1}

(f^-1)(1) = 1/{e^(e+16) +1}.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since the requestof the problem  is to calculate the derivative of the inverse of f(x), we'll conclude that the given function is bijective.

According to the rule, a bijective function is invertible.

We know that the product between the derivative of a function and the derivative of the inverse of the function is -1.

So, we'll write mathematically the statement:

[f(x)]'*[(f(x))^-1]' = 1

We infer that (f^-1)' = 1/[f(x)]'

We'll differentiate the function f(x):

f'(x) = (e^x + x + 15)'

f'(x) = e^x + 1

So,

(f^-1)' (x) = 1 / (e^x + 1)

Now, we'll calculate (f^-1)' (1):

(f^-1)' (1) = 1 / (e^1 + 1)

(f^-1)' (1) = 1/(e+1)

(f^-1)' (1) = 0.264550...