Let f(x) = e^x-1 + 5ln(x) +3x. You may assume f(x) is one-to-one. What is f^-1(4)?

Question

Let f(x) = e^x-1 + 5ln(x) +3x. You may assume f(x) is one-to-one.

What is f^-1(4)?

f^-1(4)= ?

Answer 1

Usually to find f^-1(x), you swap x and y and solve for y. However this can't be easily done for this equation.

So, we could try Newton's method

g(x) = e^x-1+5ln(x)+3x-4

g'(x) = e^x-1+5/x+3

Let x₀ = 1

x₁ = x₀ - g(x)/g'(x) = x₀ -(e^x-1+5ln(x)+3x-4)/(e^x-1+5/x+3)

= 1 -(1+0+3-4)/(1+5+3) = 1

No need to iterate further, the answer is x=1

So, f^-1(4) = 1

Answer 2

f(x) = y = e^x-1 + 5ln(x) + 3x

To find f^-1(x)

We take x = e^y-1 + 5ln(y) + 3y and solve for y, which is a very difficult task.

To find f^-1(4) we just need to solve

x = 4 = e^y-1 + 5ln(y) + 3y

If we try y = 1, we get our answer:

e^1-1 + 5ln(1) + 3(1) = e⁰ + 5(0) + 3(1) = 1 + 3 = 4

Solved by the often underestimated guess-and-check method!

Answer

f^-1(4) = 1

Answer 3

If f(a) = b, then f^-1(b) = a.

f(x) = e^x-1 + 5ln(x) + 3x

f(1) = e⁰ + 5ln(1) + 3 = 1 + 0 + 3 = 4

Since f(1) = 4, f^-1(4) = 1.