What is the equation of the line that is normal to #f(x)= e^(2x-2) /sqrt( 2x-2) # at # x=3 #?

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Answer 1 · 2018-07-04T15:30:54

#y=-8/(7*e^4)x+e^4/2+24/(7*e^4)#

Writing your function in the form

#f(x)=e^(2x-2)(2x-2)^(-1/2)#
then we can use the product and the chain rule, doing this we get

#f'(x)=2e^(2x-2)(2x-2)^(-1/2)+e^(2x-2)(2x-2)^(-3/2)*(-1/2)*2#
so we get

#f(3)=e^4-e^4/8=7/8e^4#

now we can find the slope of the normal line

#m_N=-8/(7e^4)#

now we get

#f(3)=e^4/2#
so we get

#y=-8/(7e^4)x+e^4/2+24/(7e^4)#