Tangent line, math wrong somewhere

I have an equation $\displaystyle f(x) = \frac{5x}{(3-5x)^5} $ and I want the equation of the tangent line at x = 2. I know what to do, but my math is wrong somewhere down the line.

Subbing in 2 $\displaystyle f(2) = \frac{5(2)}{(3-5(2))^5} = -0.000595 $

Taking derivative $\displaystyle f'(x) = \frac {100x+15}{(3-5x)^6} $ and then plugging in 2 $\displaystyle f'(2) = \frac {100(2)+15}{(3-5(2))^6} = 0.00183 $

So, I use equation of a line now that I have the solutions for f(x) and m.... $\displaystyle -0.000595 = 0.00183(2) + b $ and I wind up with $\displaystyle b = 0.0043 $

I get the equation of the tangent line should be $\displaystyle y = 0.00183x + 0.0043 $ ,but this is incorrect, and I was hoping somebody could eyeball this if they get a chance because I think I'm going in circles trying to fix it.

Re: Tangent line, math wrong somewhere

Your *y*-intercept is wrong. Using your decimal approximations, you should have:

*b* = (-2)0.00183 + 0.000595 = -0.003065

Re: Tangent line, math wrong somewhere

Of course, there's no reason why you should be using decimal approximations anyway...

Re: Tangent line, math wrong somewhere

I see how your math works, but why did you change 2 to a -2?

Re: Tangent line, math wrong somewhere

Quote:

Originally Posted by

**Prove It** Of course, there's no reason why you should be using decimal approximations anyway...

True.

Re: Tangent line, math wrong somewhere

Quote:

Originally Posted by

**AZach** I see how your math works, but why did you change 2 to a -2?

Think of the point-slope formula for a line.

Also, I agree with the above, I would use the true rational values...because we can. :)