i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?

Printable View

- Nov 29th 2009, 03:45 PMjohnnyc418finding the slope of the tangent line
i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?

- Nov 29th 2009, 05:36 PMmamen
- Nov 29th 2009, 06:32 PMjohnnyc418
just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do

y'=-e^-1 +2? - Nov 29th 2009, 06:38 PMJameson
If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?

Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: $\displaystyle y-y_0=m(x-x_0)$ to find the tangent line. - Nov 29th 2009, 07:01 PMjohnnyc418
- Nov 30th 2009, 12:16 AMJameson
- Nov 30th 2009, 01:52 AMjohnnyc418
- Nov 30th 2009, 01:55 AMJameson
- Nov 30th 2009, 02:14 AMjohnnyc418
so would the derivative be y'=-e^-x + (2)e^(2x-1)?

- Nov 30th 2009, 02:17 AMJameson