# Calculus Problem of the Week August 26, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution.)

The graph of a function $f(x)$ is a parabola with a vertex at $(2,3)$. Find $f'(2)$.

See the solution.

Think about any parabola (and if you can’t picture one, graph $y=x^2$ on your calculator or wolfram alpha to remember!). The vertex of any parabola is basically a turning point. That is, the graph changes from increasing to decreasing and vice-versa.

What does this have to do with derivatives? Well, the derivative at a point where the graph of a continuous function is changing from increasing (where the derivative is positive) to decreasing (where the derivative is negative) will be zero. In this example, since the vertex is at $x=2$, $f'(2)=0$.