What is the equation of the line normal to # f(x)=e^(sqrtx-x)# at # x=4#?

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Answer 1 · 2016-09-14T20:24:39

I get #y = (4e^2)/3x-(16e^2)/3+1/e^2#

#f(x) = e^(sqrtx-x)#

#f(4) = e^-2 = 1/e^2#

#f'(x) = e^(sqrtx-x)(1/(2sqrtx)-1)#

#f'(4) =e^(sqrt4-4)(1/(2sqrt4)-1) = e^(2-4)(1/4-1) = e^(-2)(-3/4) = -3/(4e^2)#

The normal line it perpendicular to the tangent, so the slope of the normal line is #-1/(f'(4)) = (4e^2)/3#.

The line through #(4,1/e^2 )# with slope #m=(4e^2)/3# is

#y-1/e^2 = (4e^2)/3(x-4)# or

#y = (4e^2)/3x-(16e^2)/3+1/e^2#