The graphs of $y=ax$ and $y=\arctan(bx)$ intersect at three distinct points if?

Question

I have a question relating to calculus and inverse trigonometric functions. Any help is appreciated.

The question is:

The graphs of $y=ax$ and $y=\arctan(bx)$ intersect at three distinct points if?

A: $0<b<a$

B: $a<b<0$

C: $a=b$

D: $b<a<0$

The correct answer is D. Upon inserting the graphs into desmos I can visually see how they intersect at three distinct points but I am struggling to understand why.

Thank you for sharing your knowledge!

Answer 1

Note that: $$\frac{d}{dx}(tan^{-1}(bx))|_{x=0} = \bigg[\frac{1}{1+(bx)^2}\cdot b\bigg]\bigg|_{x=0} = b$$ and $$\frac{d}{dx}(ax)|_{x=0} = a.$$ We want the slope of the tangent line at the origin to be steeper for the arctan function so that the other line intersects it in 3 points instead of 1 (recall the graph of artan is just an infinitely extended, bounded "S" shape).