Can you find the axis of symmetry of the graph of y = arctan(x) + arctan(6 - x)

We can start by finding the vertex. So when does dy/dx=0?
dy/dx = 1/(1+x^2) – 1/(1+(6-x)^2) = 0. Note that when x=3, this equation is true. So, the line of symmetry occurs at x=3.

Here are two assumptions I made:
1) This graph is parabolic, and therefore has only one vertex.
2) This graph is symmetric. This could also be proved using derivatives.

As a matter of curiosity can you 2 tell me if analytic geometry is still popularly taught as an independent college level course?

@dxs Nicely done, but it can be done without calculus. No assumptions are required. Refer to the hints.

@stanleybmanly , I am not knowledgeable about what is being taught these days. My teaching experience is limited to a course I taught for a semester many years ago as adjunct faculty at a community college, that was called pre-Calculus. The course included some analytic geometry. I gave a quiz at the beginning to determine what knowledge the students had of math. The students’ abilities varied considerably, making it difficult to teach the course.

@stanleybmanly I don’t know, but there are some geometry classes taught. There’s a 100-level geometry class, and a 400-level “survey of geometry” class. I’m taking that one next semester.
@LostInParadise But Calculus is so much fun. I only assumed it was parabolic because if not, it’d be a trick question. So I took that for granted. You can also argue that arctan is only a function if you make it a function of only one of the tangent shoots. As for the second one, well it again goes with your question.

What I had in mind is this.
Let f(x) = arctan(x) + arctan(6-x)
Note that h(x) = 6 -x is self-inverse.
We can make use of this by evaluating f(6 – x) = arctan(6-x) + arctan(6 – (6-x)) = arctan(6-x) + arctan(x), but this is the same as f(x).

So we have f(6-x) = f(x).

f(0)=f(6), f(1)=f(5), f(2)=f(4).
In each pair, we have values that add to 6 and are therefore equidistant from 3.
More formally, for each point (x,y) on the curve there is a point (6 – x,y) on the curve. And the midpoint of these two points is (x + (6-x))/2, y) = (3,y) That makes the line x=3 a line of symmetry.

We did not use any of the properties of arctan, and in fact we could have used for f(x) any function that treated x and (6-x) in the same way, We could have something like f(x)=x^4 + (6-x)^4 + cos(x) + cos(6-x) + 7. The above argument still holds and we would still have a line of symmetry at x=3.

We could replace 6 with any other number k and have k/2 as the axis of symmetry. For example, if we have a quadratic equation y = x^2 – bx + c, we can rewrite it as y = -x(b-x) + c, and we have a way of showing that x=b/2 is the axis of symmetry without using calculus or completing the square.

What does self-inverse mean?
Anyways, an interesting walkthrough.

The inverse of a function f is a function g such that g(f(x)) = x. For example, the inverse of f(x)=2x is g(x)= ½ x. A function f(x) is self-inverse if the inverse of f is also f. For f(x)= 6-x, f(f(x)) = x.

Oh I see. I should’ve thought on that more because it seems self-defining.
We have the slope to blame in this case then, since 1 is its own inverse.

Funny. I brought this problem up with some fellow math geeks, and they both went right to derivatives as well. I think we’re just partial to calculus.

Maybe the equation is too simple. If we add some x^4 terms, maybe people will think about some other approach.