# Why not an Absolute maximum in an open interval?

The function $x^3+x^2\: \text{has a maximun value at}\: x=-\frac{2}{3} \text{in (-1, 0) }.$
My question is why call it a Local Maximun and not an Absolute Maximum when it is the highest value in that interval anyway?

• Perhaps the point to make here is that for differentiable functions, it's often easy to establish that $x = a$ is a local max: you show that $f'(a) = 0$ and $f''(a) < 0$. Showing it's a global max may take considerably more work, so doing this first step can be useful. If there are finitely many local maxima on a closed interval, then one of them is the global max. On an open interval...you also have to check the values near the endpoints, and there may be no global max at all, as in $x^2 + x/4$ on the interval $0 < x < 1$. – John Hughes Jun 24 '15 at 12:26