What is the equation of the normal line of #f(x)=xlnx-e^-x# at #x=2#?

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Answer 1 · 2016-03-26T06:58:21

#0.5469x+y=2.34477#

As #f(x)=xlnx-e^-x#, at #x=2#, we have

#f(2)=2ln2-e^-2=2xx0.69315-0.13533=1.25097#

Hence normal passes through #(2,1.25097)#

As #f(x)=xlnx-e^-x#, #f'(x)# is given by

#(1xxlnx+x xx1/x)-e^(-x)xx(-1)# or

#(lnx+1)+e^(-x)# and at #x=2#, slope of tangent would be

#f'(2)=(ln2+1)+e^(-2)=0.69315+1+0.13533=1.82848#

Hence slope of normal would be #-1/1.82848=-0.5469#

Hence equation of normal would be

#(y-1.25097)=-0.5469(x-2)# or

#y-1.25097=-0.5469x+2xx0.5469# or

#0.5469x+y=1.0938+1.25097# or

#0.5469x+y=2.34477#