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

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

C: $$a=b$$

D: $$b

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!

• These solutions are not going to be elementary, so you're not going to be able to reasonably construct them in terms of $a$ and $b$. Do you know other tools for counting roots, like intermediate value theorem and/or mean value theorem (or Rolle's theorem)? May 26 at 11:55
• The graphs will intersect in three distinct points as long as b< a. May 26 at 11:59
• @GeorgeIvey Thank you! Not sure if the explanation will be too complicated but why is that so? May 26 at 12:01
• @TheoBendit Hmm unfortunately not, closest thing I can think of is to do with polynomial roots, thanks a lot anyways! May 26 at 12:02
• First thing is that both a and b should be of the same sign for the 3 roots condition to hold ( else there will be only 1 root )..secondly the relation between a and b can be obtained by comparing the slope of both the curves at origin
– user1012971
May 26 at 12:58

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).