What is the equation of the line normal to # f(x)=e^(2x)/(e^x+4)# at # x=2#?

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Answer 1 · 2015-12-03T16:04:57

In point slope form it is
#y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)#.

For #f(x)=e^(2x)/(e^x+4)# at #x=2#, we get #y=e^4/(e^2+4)#.

The slope of the tangent line is given by:

#f'(x) = ((2e^(2x))(e^x+4)-e^(2x)(e^x))/(e^x+4)^2 = (e^(3x)+8e^2x)/(e^x+4)^2#.

At the point where #x=2#, the slope of the tangent is

#m_(tan) = (e^6+8e^4)/(e^2+4)^2#.

So the slope of the normal line is

#m_(norm) = (e^2+4)^2/ (e^6+8e^4)#.

The equation of the line (in point slope form) through #(2,e^4/(e^2+4))# with slope #m = (e^2+4)^2/ (e^6+8e^4)# is

#y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)#.

Rewrite using algebra as you see fit. (Or as your grader demands.)