Increasing / Decreasing Functions

Students who are new to calculus might immediately seek to take a derivative (and rightly so!) because the derivative gives insight about whether a function is increasing or decreasing. Indeed, the derivative of \(y = x^3\) is \(y' = 3x^2\), which takes on a value of 0 at \(x = 0.\) But does this mean the function isn't always increasing?

Before we start answering this question from the perspective of calculus, we should actually know what it means for a function to be increasing. We say a function is increasing on the interval \(I\) if \(f(a) \leq f(b)\) for all \(a < b\) in \(I.\) Indeed, we have that for every choice of \(a < b\) on the real number line, it is guaranteed that \(a^3 \leq b^3\), that is \(f(a) \leq f(b).\) This is true even if one of \(a\) or \(b\) is 0! Therefore, it is \(\boxed{\textbf{true}}\) that the function \(y = x^3\) is always increasing.

So now we must defuse the argument that says, "We know that \(y'(0) = 0\), which implies that the function \(y = x^3\) is not increasing at \(x = 0\)." Simple enough. This argument is a misrepresentation of what the derivative of a function can tell us. The derivative merely tells us that if \(f'(x) > 0\) for all \(x\) on a given interval \(I\), then \(f\) is increasing on \(I\). That does not imply that \(f\) is not increasing if \(f'(x) = 0\) for a particular \(x\) in \(I\).

Taking the derivative of \(f(x)\) gives \( f'(x)= e^x(x^2+(a+2)x+3a) \). If \(f(x)\) is strictly increasing, then \(f'(x) > 0\), or

\[\begin{align} e^x(x^2+(a+2)x+3a) & > 0\\ x^2 +(a+2)x+3a & > 0. \qquad(1) \end{align}\]

Since this equation has no real roots, it must be the case that the discriminant is negative, i.e.,

\[\begin{align} (a+2)^2-12a & < 0\\ a^2-8a+4 & < 0. \end{align}\]

This is satisfied for integers \(a\) in the interval \( ( 4-2\sqrt{3},4+2\sqrt{3})\), or the integers \(a=1, 2, \ldots 7\). These values also satisfy equation (1) for all \(x\), implying the number of possible integer values \(a\) is \(7\).