Slope of normal to a given function's inverse

[Elixer]

Homework Statement

f(x) = 2x² + 4e^(5x)
is invertible. Give the slope of the normal line to the graph of f^-1 at x = 4.

Homework Equations

(Given in question)

The Attempt at a Solution

I don't know how to solve this question. But , I found the following:-

f(4) = 32 + 4e²⁰

let y = f^-1(x), then
dy/dx = 1/f ' (y)
f ' (x) = 4x + 20e^(5x)
Hence, dy / dx = 1/(4y + 20e^(5y))
I don't know if I am heading in the right direction.
Please help!
Thank you.

 

[LCKurtz]

Homework Statement

f(x) = 2x² + 4e^(5x)
is invertible. Give the slope of the normal line to the graph of f^-1 at x = 4.

Actually,the function is not invertible because it isn't 1-1. But you might observe that f(0) = 4 and it is invertible in a neighborhood of x = 0.

Homework Equations

(Given in question)

The Attempt at a Solution

I don't know how to solve this question. But , I found the following:-

f(4) = 32 + 4e²⁰

But f(4) isn't relevant to the question. The question is about the inverse function's slope at x = 4. Remember that f(0) = 4 means (0,4) is on the graph of y = f(x) and it also means that f^-1(4) = 0 and (4,0) is on the graph of y = f^-1(x).

The slope of that function at that point is what you are looking for. Does that help?

 

[Elixer]

But f(4) isn't relevant to the question. The question is about the inverse function's slope at x = 4. Remember that f(0) = 4 means (0,4) is on the graph of y = f(x) and it also means that f^-1(4) = 0 and (4,0) is on the graph of y = f^-1(x).

The slope of that function at that point is what you are looking for. Does that help?

So, (4,0) lies on the graph of f^-1(x),
y = f^-1(x)
dy / dx = 1/(4y + 20e^(5y))
Hence , slope of f^-1(x) , dy / dx = 1/20
which implies , slope of the normal to the graph = -1 /(1/20) = -20
Ans = -20

Am I correct?
Thanks a lot!

 

[LCKurtz]

So, (4,0) lies on the graph of f^-1(x),
y = f^-1(x)
dy / dx = 1/(4y + 20e^(5y))

If you calling y = f^-1(x) then I would write

y' = 1 / f'(x) = 1/(4x + 20e^(5x))

and yes, your answer looks correct.