# The sign of the times

## Problem

My friend has sketched a non-constant curve $f(x)$ which passes through the origin. She knows that its derivative exists at all points. Is it possible that $f(x)$ could satisfy

$$

f(x)\times\frac{df(x)}{dx} \leq 0

$$ for all $x$?

I'll need a very clear explanation to convince me!

My other friend has sketched a curve whose derivative exists at all points, but his does not pass through the origin. Is it possible that the same condition might hold?

Problems in calculus can often be considered from either an algebraic or geometric viewpoint and calculus is fundamental in the advanced study of geometry as well as areas of theoretical physics, such as string theory and relativity.

## Getting Started

So, the question can be rephrased as: can the sign of a curve and the sign of its gradient be opposite everywhere?

## Student Solutions

Consider curves that pass through the origin and have the required property that $$f(x)\times\frac{df(x)}{dx} \leq 0$$

Suppose $f(x)$ is positive for some positive $x=x_0$. If we require $f(0)=0$, this means the gradient must be positive for part of the interval $[0, x_0]$, which would mean that $f(x)\times\frac{df(x)}{dx} > 0$ for these points.

Similarly, if we suppose $f(x)$ is negative for some positive $x=x_0$, we find that the gradient will have to be negative for part of the interval $[0,x_0]$, which would mean that $f(x)\times\frac{df(x)}{dx} > 0$ for these points.

We therefore need $f(x)\equiv0$ for all $x\ge0$.

For $x< 0$, we can choose f so that $f(x)\not\equiv0$. For example, we can choose f to be positive but have a negative gradient for all $x< 0$, which would mean $f(x)\times\frac{df(x)}{dx} < 0$.

Let's try choosing $f=-x$ for $x< 0$:

This satisfies the condition $f(x)\times\frac{df(x)}{dx} \leq 0$ for all x. The other condition was that the derivative exists for all x. For $x> 0$, $\frac{df(x)}{dx} = 0$, and for $x< 0$, $\frac{df(x)}{dx}=1$. However, what's the derivative at $x=0$? As we approach from the right, $\frac{df(x)}{dx}=1$, but as we approach from the left, $\frac{df(x)}{dx}=0$. Therefore, our derivative doesn't exist at $x=0$ as was required, so we need to think of something else! (For a rigorous treatment of this, see a first year analysis course.)

We need to think of another function with behaviour like the previous one, but with a gradient of $0$ as we approach $x=0$. Let's try the next simplest polynomial, a quadratic: $f(x) = x^2$. This passes through the origin, and has derivative $\frac{df(x)}{dx}=2x$. So at $x=0$, the derivative is $0$ as we approach from both directions, so therefore exists. We've found a function that has the desired properties!

If we don't have the requirement that $f(0)=0$, there are lots of examples, e.g. $f(x)=exp(-x)$.

Here, $f(x)>0$ for all x, and the gradient exists for all x and is negative: $\frac{df(x)}{dx} = -exp(-x) <0,$ as required.