# Prove that a function is strictly increasing

I have the following function $$f(x)= x\left (1 + \displaystyle{\sum_{j=1}^M \ \frac{p}{1 + D_{j}\cdot x}} \right)- t.$$

$$x$$ is the only variable (the others are positive constants). We have been asked to show two things:

1. Function is positive for some large values of $$x$$.
2. Function is strictly increasing for positive $$x$$.

For (1) I can just use a few values to prove it. For the second part, I'm thinking of taking the derivative of this function and proving that it will always be positive. I got the derivative and said that for very large values of $$x$$ (as $$x\to\infty$$), the derivative will converge to $$1$$. Thus it will be strictly increasing. Just hoping someone can help me confirm this.

I got the derivative and said that for very large values of $$x$$ (as $$x\to\infty$$), the derivative will converge to $$1$$

All that this proves is that $$f(x)$$ is increasing for sufficiently large $$x$$. Since you were asked to prove that $$f(x)$$ increases at all positive values of $$x$$, it's not enough.

Here's a hint: what's the derivative of $$g(x) = \frac{x}{1+ k x}$$, where $$k$$ is a constant? Can you find a simplified expression? What can you say about its sign?

• Thank you for this. The thing you've given me would be 1/(1+kx)^2, which is always positive. Since my function is a positive value multiplied to this, we know this is always positive. Is that correct? Sep 29, 2022 at 15:17

Here is a way of doing it

For 1 you just take $$x>t$$.

2 boils down to showing that $$x\left (1 + \displaystyle{\sum_{j=1}^M \ \frac{p}{1 + D_{j}*x}} \right) when $$0. To show this, is sufficient to show that $$\frac{x}{1 + D_{j}*x}<\frac{y}{1 + D_{j}*y}$$

Rearranging the terms, we have $$x+D_j*xy