Find all functions f having the indicated property: All tangents to the graph of f pass through the origin

Equation of a tangent line to the graph of function `f` at point `(x_0,y_0)` is given by `y=f(x_0)+f'(x_0)(x-x_0).`

Since every tangent passes through the origin `(0,0)` we have

`0=f(x_0)+f'(x_0)(0-x_0)`

`x_0f'(x_0)=f(x_0)`

Let us write the equation using usual notation for differential equations.

`x (dy)/(dx)=y`

Now we separate the variables.

`(dy)/y=dx/x`

Integrating the equation, we get

`ln y=ln x+ln c`

`c` is just some constant so `ln c` is also some constant. It is only more convenient to write it this way.

Taking antilogarithm gives...

Equation of a tangent line to the graph of function `f` at point `(x_0,y_0)` is given by `y=f(x_0)+f'(x_0)(x-x_0).`

Since every tangent passes through the origin `(0,0)` we have

`0=f(x_0)+f'(x_0)(0-x_0)`

`x_0f'(x_0)=f(x_0)`

Let us write the equation using usual notation for differential equations.

`x (dy)/(dx)=y`

Now we separate the variables.

`(dy)/y=dx/x`

Integrating the equation, we get

`ln y=ln x+ln c`

`c` is just some constant so `ln c` is also some constant. It is only more convenient to write it this way.

Taking antilogarithm gives us the final result.

`y=cx`

There fore, our functions `f` have form `f(x)=cx` where `c in RR.` 

Graphically speaking these are all the lines that pass through the origin. Since the tangent to a line at any point is the line itself the required property is fulfilled.

Graph of several such functions `f` can be seen in the picture below.