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

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Answer 1 · 2016-12-04T00:58:06

The equation of the line normal to #f(x) = sqrt (e ^ sqrtx)# at #x=4# is:

#y-e =-8/e(x-4)#

The equation of the line normal to #f(x)# in #x=bar x# is given by:

#y-f(bar x) =-1/(f'(bar x))(x- barx)#

So we calculate the first derivative of #f(x)#:

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

using the chain rule:

#f'(x) = e^(1/2x^(1/2))*1/2(1/2x^(-1/2)) = e^(1/2sqrt(x))/(4sqrt(x))#

For #bar x = 4#:

#f(bar x) = e#
#f'(bar x) = e/8#

So the equation of the normal line is:

#y-e =-8/e(x-4)#