A difficulty in understanding a statement in example 10.6.6 Petrovic.

Question

The example is given below:

But I do not understand why the quadratic function can not have a local maximum when $a > 0$ (this was mentioned at the last line of the example), could anyone explain this for me please?

Answer 1

Let $f(t) = at^2 + bt + c$, with $a \gt 0$. If you use some example values for $a, b, c$ and plot this, you will see it's a concave up parabola with an axis of symmetry, and a minimum, at $t = \frac{-b}{2a}$. For example, $f\left(t\right) = t^2 - 2t + 1 = \left(t - 1\right)^2$ is a parabola with an axis of symmetry at $t = 1$ and a minimum of $f(1) = 0$.

Using calculus, this is easiest to see due to $f''(t) = 2a \gt 0$ so any extrema must be local minima.

Also, note that as $t \to \pm \infty$, the first term of $at^2$ predominates, so $f(t) \to \infty$ and there can't be a global maximum.

In addition, as $f'(t) = 2at + b$ is a linear function, there can only be at most one local extrema, so it can't be a local maximum as the continuity of $f\left(t\right)$ would require there be local minimums on either side for $f\left(t\right) \to \infty$ as $t \to \pm \infty$.