Question

What is the equation of the tangent line of `f(x) = (x^2e^(x)-x)/(x-1)` at `x=2`?

Answer 1

`y = 30.56x-33.56`. The Socratic graph is tangent-inclusive. The point of contact of the tangent is (2, 27.56).

Explanation:

`y(x-1)=x^2e^x-x`. Differentiating,

`y'(x-1)+y=2xe^x-x^2e^x-1=xe^x(2-x^2)-1`

At `x = 2, y =4e^2-2=27.56`, nearly, and

`y' = 4e^2+1=30.56`, nearly.

So, the equation to the tangent is

`y - 27.56 = 30.56(x-2)`. Simplifying,

`y = 30.56x-33.56`

graph{(y-30.56x+33.56)(y(x-1)-x^2e^x+x)=0 [-10, 10, -30, 30]}