If we have a function $f(x)>0$ that has the following property:
$$f''(x)>0, $$ where this property only holds for $x>0$. What is the angle $\theta$ that a tangent line to the function makes with the $x$-axis, given that the tangent line must pass through the origin, and the point of tangency with $f(x)$ occurs in the first quadrant?
Here is what I have done:
Let the tangent line be $y=mx$, where $m>0$
So we let the two functions equal each other, and take the derivative of both sides with respect to $x$ to give us another equation as well, since their slopes will be equal at the intersection. Now we have the two equations:
$$f(x)=mx$$ $$f'(x)=m$$
Now after solving for $m$, I obtained the following result:
$$m=f'\left(\frac{f(x)}{f'(x)}\right)$$
Which will leave us with:
$$\theta=tan^{-1}\left(f'\left(\frac{f(x)}{f'(x)}\right)\right)$$ to the $x$-axis.
Are my steps in solving this correct? Will this leave us with an angle independent of x if we plug in a function that satisfies the conditions?