# Equation of Tangent Line

1. Dec 7, 2004

### chjopl

Find the equation of the tangent line at x=2
F(X)=2x+7x(e^(x-2))

I know you put 2 in to find the y intercept and you take the derivative to find the slope. I get 2+(7e^(-2)x + 7e^-2)e^2 for my derivative and y=23x+18 as my answer and i know that is not correct.

2. Dec 7, 2004

### shmoe

You derivative looks off. Remember,

$$\frac{d}{dx}e^{x-2}=e^{x-2}\frac{d}{dx}(x-2)=e^{x-2}(1)=e^{x-2}$$

and apply the product rule carefully!

edit-ahh, Galileo is right, your slope is fine. Your posted derivative may be a casualty of ascii.

Last edited: Dec 7, 2004
3. Dec 7, 2004

### Galileo

You got the slope right. So y=23x+b. Now use the fact that the line passes through the point (2,f(2)) to find b.

4. Dec 7, 2004

### dextercioby

The derivative should be:
$$f'(x)=2+[7\exp({-2})](\exp{x}+x\exp{x)$$
Insert 2 to find the slope.It's 23.
Compute f(2).It's 18.With the last 2 numbers,u should be able to find the equation for the tangent line.

We're just like eagles when it comes to simple problems.Three almost identical posts at the same time...

Last edited: Dec 7, 2004
5. Dec 7, 2004

### chjopl

so was y=23x+18 correct?

6. Dec 7, 2004

### dextercioby

Obviously not,try to follow my post correctly.It's easy maths,after all.

7. Dec 7, 2004

### HallsofIvy

Staff Emeritus
What is 23(2)+ 18?