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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.

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2 Answers 2

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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?

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  • $\begingroup$ 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? $\endgroup$
    – pasha
    Sep 29, 2022 at 15:17
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Here is a way of doing it

For 1 you just take $x>t$.

2 boils down to showing that $$ x\left (1 + \displaystyle{\sum_{j=1}^M \ \frac{p}{1 + D_{j}*x}} \right)<y\left (1 + \displaystyle{\sum_{j=1}^M \ \frac{p}{1 + D_{j}*y}} \right) $$ when $0<x<y$. 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 $x+D_j*xy<y+D_j*xy\iff x<y.$

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