Consider the functions f(x) = ln ln where and 3

Question

x
e
( 2 ) x≥1 g(x) =
f(x)
f
−1
(x)

Find the equation of the tangent to g(x) at the point where x = 1.

Answer 1

To find the equation of the tangent to g(x) at the point x = 1, we first need to find the derivative of g(x).

We are given that g(x) = f(x) / f^-1(x).

First, let's find the derivative of f(x).

f(x) = ln(ln(x))

Using the chain rule, the derivative of f(x) is given by:

f'(x) = (1/ln(x))(1/x) = 1/(x ln(x))

Next, let's find the derivative of f^-1(x).

To find the derivative of f^-1(x), we need to solve for f^-1(x).

Since f(f^-1(x)) = x, we have:

ln(ln(f^-1(x))) = x.

Applying the natural logarithm to both sides, we get:

ln(f^-1(x)) = e^x.

Taking the exponential of both sides, we have:

f^-1(x) = e^(e^x).

Now, let's find the derivative of f^-1(x). Using the chain rule, we have:

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

Now, let's find the derivative of g(x) using the quotient rule:

g'(x) = (f'(x)f^-1(x) - f(x)(f^-1(x))') / (f^-1(x))^2

Plugging in the values we found earlier, we have:

g'(x) = (1/(x ln(x)))(e^x * f^-1(x)) - ln(ln(x))(e^x * f^-1(x))/(f^-1(x))^2

Simplifying, we have:

g'(x) = [(e^x)/(x ln(x))] - [(ln(ln(x))(e^x))/(f^-1(x))]

Now, let's evaluate g'(1) to find the slope of the tangent line:

g'(1) = [(e^1)/(1 ln(1))] - [(ln(ln(1))(e^1))/(f^-1(1))]

Since ln(1) = 0, we have:

g'(1) = e - 0 = e

The slope of the tangent line is e.

Now, let's find the y-coordinate of the point of tangency. Plugging x = 1 into g(x), we have:

g(1) = (ln(ln(1)))/(f^-1(1))

Since ln(1) = 0, we have:

g(1) = 0

The y-coordinate of the point of tangency is 0.

Therefore, the equation of the tangent line to g(x) at the point where x = 1 is:

y = ex.