Question 1192491
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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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<pre>
You correctly determined the derivative function  {{{(dy)/(dx)}}} = -{{{1/e^x}}}.


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^(-x)}}}

at the point x= 2.


        So, the slope is  m = -{{{1/e^2}}}.


Now you know the point on the plot of the function: it is  ({{{x[0]}}},{{{y[0]}}}) = ({{{2}}},{{{1/e^2}}});

and you know the slope:  it is  m = -{{{1/e^2}}}.


So, the equation of the tangent line is  {{{y - y[0]}}} = {{{m*(x-x[0])}}},  or

    y - {{{1/e^2}}} = -{{{(1/e^2)*(x - 2)}}}.


It is your <U>ANSWER</U>.


You can also use any other EQUIVALENT form of the last equation.
</pre>

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Solved and explained.


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