Question

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

Answer 1

`0.5469x+y=2.34477`

Explanation:

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`