Question

What is the equation of the tangent line of `f(x) =(x-1)(3x-3e^(x^2-1))` at `x=1`?

Answer 1

`y=0` -( the `x` axis)

Explanation:

For the given function `f(1) = 0`. So, the tangent line that we want goes through `(1,0)` and has a slope given by `f prime (1)` .

To find the derivative of `f(x)` quickly, note that according to the formula `d/dx (uv) = u {dv}/dx +v {du}/dx`, we do not need to differentiate the second factor of our expression - since that will be multiplied by `(x-1)` and hence will vanish at `x=1`. So

`f prime (1) = 1 \times (3 times 1-3 times exp(1^2-1))=3-3 = 0`

Thus the tangent is `y = 0` - the `x` axis.