SOLUTION: Find the equation of the tangent to; f(x) = e^-x at the point where x=2. So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation and

Algebra ->  Functions -> SOLUTION: Find the equation of the tangent to; f(x) = e^-x at the point where x=2. So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation and       Log On


   



Question 1192491: Find the equation of the tangent to;
f(x) = e^-x at the point where x=2.
So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation and find the y value.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the equation of the tangent to;
f(x) = e^-x at the point where x=2.
So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation
and find the y value.
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You correctly determined the derivative function  %28dy%29%2F%28dx%29 = -1%2Fe%5Ex.


To proceed further, you should substitute x= 2 into the formula. 

Doing this way, you will find the value of the derivative at the point x= 2, which is the slope of the function

    f(x) = e%5E%28-x%29

at the point x= 2.


        So, the slope is  m = -1%2Fe%5E2.


Now you know the point on the plot of the function: it is  (x%5B0%5D,y%5B0%5D) = (2,1%2Fe%5E2);

and you know the slope:  it is  m = -1%2Fe%5E2.


So, the equation of the tangent line is  y+-+y%5B0%5D = m%2A%28x-x%5B0%5D%29,  or

    y - 1%2Fe%5E2 = -%281%2Fe%5E2%29%2A%28x+-+2%29.


It is your ANSWER.


You can also use any other EQUIVALENT form of the last equation.

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Solved and explained.

If you have question/questions, then let me know.


If you will post your questions, please refer to the problem's ID number 1192491 .