# Find point $P$ so that the tangent to the function plot passes through origin

Find the point $P$ situated on the plot of the function $f(x) = e^x + x$, where the tangent to the plot passes through the origin.

Here is my approach:

Let $d: y - f(x_0) = f'(x_0)(x - x_0)$ be the equation of the tangent to the function plot at point $M(x_0, f(x_0))$.

The tangent passes through the origin, which means that $y = xf'(0)$.

$f'(x) = e^x + 1$, we get $f'(0) = 2$.

So, the equation of the tangent to the function plot which passes through the origin is $d: y - 2x = 0$.

Now, we need to find the intersection of the function plot with this tangent line. From the equation of the tangent we get $y = 2x$, so $f(x) = 2x$.

Plugging what we've just got into $f(x) = e^x + x$, we get $e^x - x = 0$, which has no real solutions.

So, the point P doesn't exist.

Is the solution I provided a correct one? Are there any mistakes?

It took me a while to understand what was going through your mind, but I think I got it now. The whole problem comes from

The tangent passes through the origin, which means that $y=xf′(0)$.

where it seems you're forcing the plot of f to pass through the origin, instead of the tangent. Going back to the tangent equation

$$y = f(x_0) + f'(x_0)(x - x_0)$$

We must make $(0, 0)$ a point of this rect, so

$$0 = f(x_0) + f'(x_0)(0 - x_0) \implies x_0 = \frac{f(x_0)}{f'(x_0)}$$

assuming nothing crashes when dividing. Now, should you a find such an $x_0$ then $(x_0, f(x_0))$ is the point in the plot you're looking for.

HINT:

$$\dfrac{y-f(x_0)}{x-x_0}=f'(x_0)$$

$$\implies\dfrac{y-(e^{x_0}+x_0)}{x-x_0}=e^{x_0}+1$$

Now this needs to pass through the origin.

Can you find $x_0(=1)$ and hence $f(x_0)$ from here?