1
$\begingroup$

On first look, doesn't any number suffice? For example, if I let $n=1$ then $1+\frac{1}{10} < \alpha < 1+\frac{2}{10}$ is valid right? Tho, I think the question is asking in the context of the rest of the question? Am at part iv

enter image description here

I don't really get the reasoning behind the answer tho

enter image description here

How does the pieces fall into place? Why the expression for $g(x)$. How do I get the seemingly arbitrary numbers 1.7 and 1.8?

1 Answers 1

1

The expression for $g(x)$ comes from (iii). Apparently $\alpha \ln \alpha - \ln(1+\alpha) = 0$, so $g(\alpha) = 0$. As for how to get the numbers 1.7 and 1.8: The question sort of indicates that you'll probably find something fairly near 1, so I guess just trying values $g(1+\frac{n}{10})$ for $n$ from $0$ to $10$ until you get a change of sign could be expected to work. There's probably a better way.

As for your first question, yes, of course it's in the context of the rest of the question. Otherwise what is $\alpha$? How could you claim that any $n$ works, without knowing what $\alpha$ is?