Question

What is the equation of the line tangent to `f(x)=ln(x^2+x)*(x^2+x)` at `x=-1`?

Answer 1

The tangent line cannot be found in the ordinary sense because the function is not defined for `x=-1`.

However, we can see that for `x->-1` the tangent line to the curve approaches the vertical line `x=-1`

Explanation:

In general, the equation of the tangent line to the curve `y=f(x)` in the point `(bar x, f(barx))` is given by:

`y(x) = f(barx) + f'(barx)(x-barx)`

Now, we can observe that `f(x)` is only defined for `(x^2+x) > 0`,
which means for `x in (-oo,-1) uu (0,+oo)`, so strictly speaking it is not defined in `x=-1`.

However, remembering that:

`lim_(t->0) tlnt=0`

we can remove the discontinuity and pose `f(-1) = 0`.

We now calculate the first derivative:

`f'(x) = d/(dx) [(x^2+x)ln(x^2+x)]`

`f'(x) = (x^2+x) *d/(dx) [ln(x^2+x)]+ d/(dx)[(x^2+x)]ln(x^2+x)`
`f'(x) = (x^2+x) (2x+1)/(x^2+x)+ (2x+1)ln(x^2+x)`
and finally:

`f'(x) = (2x+1)(1+ln(x^2+x))`

We can see that for `x->-1^-` (from the left, as `f(x)` is not defined for `-1 < x < 0`)

`lim_(x->-1^-) f'(x) = +oo`.

So the derivative cannot be extended for continuity and we cannot use the formula in the standard form, but we can still proceed in a different way.

Let us consider the normal to the tangent line to the curve for values of `x` approaching `x=-1` from the left in succession:

`barx_n = -1-1/n`

These are the family of lines:

`y_n(x) = f(barx_n) -1/(f'(barx_n))(x-barx_n)=-x/(f'(barx_n))+(f(barx_n)+barx_n/(f'(barx_n)))`

We can easily persuade that for `n->oo` these lines approach to the line:

`y=0`

and in the same way the tangent lines approaches the vertical line:

`x=-1`