I am being told to find the intervals on which the function is increasing or decreasing.
It is a normal positive parabola with the vertex at $(3,0).$ The equation could be $y = (x-3)^2,$ but my confusion comes from the interval on which the parabola is increasing: I would think increasing is $(3,\infty)$ and decreasing is $(-\infty, 3)$.
However the text book teaches to use $[3,\infty)$ and $(-\infty, 3].$ Can you explain this. I thought the function was constant at $x=3$.
Thanks
Best Answer
I like this question, because it touches upon a subtle point.
First, let me say that it would be better to write "the function is flat at $x=3$" than "the function is constant at $x=3$". Any function is constant at any one point, if you look at it that way. Even better than "the function is flat" might be "the function has a horizontal tangent at $x=3$". But anyway.
I agree with the book, when it includes the endpoints of the intervals. Because what is the definition of being increasing on some interval $I$? It is that if $x_1<x_2$ then $f(x_1)<f(x_2)$ for all $x_1,x_2\in I$. With this definition, you see that the function is actually increasing on $[3,\infty)$, and not just on $(3,\infty)$. Indeed, if $x_1=3$, then however I choose $x_2>3$, I will have that $f(x_1)<f(x_2)$. The definition holds true, even if I include the endpoint.