How to show this function is increasing both intuitively and using formula

Question

In a book, it is claimed that the following function is obviously increasing as $x$ gets larger. I tried a few numerical examples for $x$, and it appears to be so. However, it is not clear to me how come it is so obvious. Could anyone point out the reason, please?

$g(x) := \frac{(65-x)*0.8^{65-x}}{1-0.8^{65-x}}; 0 < x < 65.$

My try is as below, let $x_1 < x_2$, it is only necessary to show that $g(x_1) / g(x_2) <1$. Hence,

$ \begin{align} \frac{g(x_1)}{g(x_2)} = \frac{65-x_1}{65-x_2} * 0.8^{x_2-x_1}*\frac{1-0.8^{65-x_2}}{1-0.8^{65-x_1}}. \end{align} $

It is not hard to see the first fraction term is greater than 1; the second term less than 1; and the third item also less than 1. However, it is not clear why the product of these three is less than 1. I guess this relates to the order of increase; i.e. linear increase/decrease is slower than exponential. However, I could not finish the above formula to reach the desired answer.

Moreover, is there any intuitive explanation why $g(x)$ is obviously increasing, please? Thanks.

Update:

I think Rhys' argument can explain when $x$ is small, but not when $x$ gets closer to the upper boundary. The plot of numerator and dominator as well as the function itself is as below. So it is still not clear to me why $g(x)$ is increasing when $x$ is close to 65.

Answer 1

It's "obvious" (I don't exactly agree as it took a minute to get my head round it) because the numerator of $g(x)$ is increasing; consider that if you have $$(\frac 45)^{30}\cdot 30=(\frac45)^{29}\cdot 24 $$ (when $x=35$), which is obviously smaller than the $(\frac45)^{29}\cdot 29$ (when $x=36$)

and its denominator is decreasing for the same reasons. Because of this, the function is increasing.

Answer 2

I don't find it obvious, either. You have identified the problem in the $65-x$ term. The graphs Alpha shows strongly support the increasing claim. The claim that exponential dominates over linear is true as the exponent goes to infinity, but we are not doing that here. For $x$ small the numerator is increasing because each increase of $1$ increases the $0.8^{65-x}$ term by a factor $1.25$ and decreases the $65-x$ term by $1$. When the $65-x$ term is large that is increasing, but when $x$ goes from $63$ to $64$ the numerator decreases from $1.28$ to $0.8$. The denominator decreases enough so the expression still increases, but this supports the claim that it is not obvious that the expression is increasing.