# Slope of the normal line

1. Jul 10, 2013

### Jude075

1. The problem statement, all variables and given/known data
The function f(x)=5x2+3e2x is invertible. Give the slope of the normal line to the graph of f-1 at x=3

2. Relevant equations

(f-1)'(x)=1/f'(f-1(x))
The slope of the normal line × the slope of the tangent =-1
So basically I just need to find -f'(f-1(x)) at x=3

3. The attempt at a solution
Derivative of the original function is 10x+6e2x, but I cannot find f-1(x) as I am not able to inverse the function.

2. Jul 10, 2013

### Millennial

You don't need the full expression of the inverse to find its value at a particular point. The value $f^{-1}(3) = x_0$ is supposed to satisfy the equation $\displaystyle f(x_0) = 5x_0^2 + 3e^{2x_0} = 3$. Any idea what that $x_0$ might be?

3. Jul 10, 2013

### Jude075

This is just as difficult as finding the full expression for me. Please elaborate it more. Thank you!

4. Jul 10, 2013

### micromass

Staff Emeritus
There's no way to solve it analytically. The only thing to do is to guess some numbers and see if they work. Luckily, there is some easy number that does the trick!

5. Jul 10, 2013

### Millennial

Micromass is correct, if you waste time trying to find the inverse in a test, you will lose time for nothing. Here are a few tips for you to find the obvious solution to that equation:

- $x^2$ is not a nonzero function.
- $e^x$ is a nonzero function.
- The coefficient of $e^{2x_0}$ in the expression is the value of the expression.

6. Jul 10, 2013

### Mentallic

We want to find the value of x such that

$$5x^2+3e^{2x}=3$$

Usually we would go about solving for x, but in this case, we can't because there is no analytic solution. That is, we don't have any easy way of writing x = ...

But the question purposely used the value of 3, which gives us a nice easy value for x that we can guess and find. If the question gave us a 4 or various other numbers instead, then we'd be out of luck and would have to go to the trouble of approximating a value for x.

So, what value of x solves that equation? Notice that x2 is never negative, so 5x2 is never negative, and e2x is always more than zero hence 3e2x>0. Hence since we are adding these two non-negative expressions, and we want them to add up to just 3, we should probably look at small values of x (not large negative else x2 is a large value).

7. Jul 10, 2013

### Jude075

It's 0,so the slope is -6 :)
Thank you guys very much!

8. Jul 10, 2013

### Mentallic

That's it

9. Jul 10, 2013

### skiller

Sorry to bump this thread but I was just reading it and I had a different approach and came up with a different answer and I can't see where my logic has gone wrong. I'm pretty tired at the moment so I'm sure someone will put me right here...

The function is:

$$f(x) = 5x^2 + 3e^{2x}$$
The question is:

What is the slope of the normal of $f^{-1}$ at $x=3$ ?

the slope of the normal = -1/the slope of tangent

Also, as the graph of $f^{-1}$ is just the graph of $f$ reflected in the line $y=x$, then surely...

the slope of the tangent of $f^{-1}$ = 1/the slope of tangent of $f$

Therefore, the slope of the normal of $f^{-1}$ = $-f'$

$$f'(x) = 10x + 6e^{2x}$$
$$-f'(3) = -30 - 6e^6$$
So the answer to the question is: $-6(e^6+5)$

Where is the error in my logic?

Last edited: Jul 10, 2013
10. Jul 10, 2013

### Jude075

You need to plug in f-1(x) instead of x, which is 0 .

11. Jul 11, 2013

### skiller

Got it now, sorry.

Told you I was tired! :zzz: