As the title states, Determine the equation of the tangent to the curve $y = x\ln(x)$ that passes through the point $(0, −𝑒)$
I've solved for the first derivative with respect to $x$, to determine the gradient of the tangent, which I found to be $\frac{dy}{dx} = \ln(x) +1$ .
However, when I sub in the point $x = 0$ the function is undefined, as $\ln(0)$ is undefined. Could someone please provide an alternative approach.
Best Answer
Hint:
As the comment from saulspatz states, the point $(0, -e)$ is not the point of tangency. The point is not on the curve but on the line. Here is an approach. Take a point $P \ (x_0, x_0 \ln x_0)$ on the curve. The tangent line at point $P$ passes through point $Q \ (0, -e)$. So find slope of $PQ$ given two points. Now as this line is tangent to the curve at $P$, its slope should also be equal to $y' = 1 + \ln (x_0)$. Equating them should give you $x_0$ and you should then easily get the equation of the line which is $y+e = (1 + \ln (x_0)) \ x$