Let g be a differentiable function of domain $[0,10]$ such that $g(5-x)=g(5+x), \forall x\in [0,5]$ and $g''(x) \geq 0, \forall x\in[0,10]$. Show that the slope of the tangent line to the graph of g in the point of x-coordinate 5 is zero and the function has a minimum in that point.
For this, I know that if the second derivative is always non-negative, the graph is always concave upwards, so it must have a minimum. Also, it should be a simple parabola symmetric around $x=5$, because for $x=0$ we have the obvious $g(5-0) = g(5+0) \Leftrightarrow g(5) = g(5)$. Because I have a single parabola which is concave upwards, I have a function that is first decreasing to a minimum and then increasing. Therefore, I am reasoning that then I should have $g'(5-x)=-g'(5+x)$. Can I make this point here? How do I go on from here? Do I have to bring the Corollary of Intermediate Value Theorem to prove that the differential function has at least a zero? How do I formalize the whole proof?
Thank you very much for your help.
All the best!
Best Answer
Yes, we have $g'(5-x)=-g'(5+x), \forall x\in [0,5]$. With $x=0$ we get $g'(5)=-g'(5)$, hence $g'(5)=0$.
Furthermore: $g''(5) \ge 0.$ Conclusion ?