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Given that $1<a<2$ and $1<b<2$

I have to find the value of $x$ and $y$ such that $y-x=0.15$ for $x<\frac{a+b}{2a+b}<y$

$\frac{1}{8}<\frac{1}{2a+b}<\frac{1}{3}$ and $2<a+b<4$ so $\frac{1}{4}<\frac{a+b}{2a+b}<\frac{4}{3}$

but $\frac{4}{3}-\frac{1}{4}=\frac{5}{12}=0.416667$

we see that $0.416667>0.15$

so I need to use another form of $f(x)=\frac{a+b}{2a+b}$

I heared about reduced form of $\frac{ax+b}{bx+c}$ I think I have to use it but I don't know how to do ... .Can somone help me ?

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  • $\begingroup$ The use of the same symbol $x$ here for a lower bound of the expression $(a + b)/(2a + b)$, and in the expression "$f(x)$" (where a value for "$x$" has not been defined) is unfortunate - and in my answer, I may not have helped matters, by presuming on your behalf to give, retrospectively, a value to the "$x$" in your "$f(x$)"! Sorry about that. (And welcome to MathSE, by the way - I'm quite new here myself - I hope this hasn't been too bumpy a ride so far!) $\endgroup$
    – Calum Gilhooley
    Commented Mar 5, 2015 at 22:15
  • $\begingroup$ Im also new. and english is not my mother language I can't understand all things but I can know what is meaning . one thing is that i didnt understand what are u meaning by express the fraction as a function of the single quantity x=b/a ? $\endgroup$
    – juste smile
    Commented Mar 5, 2015 at 22:19
  • $\begingroup$ I meant: divide both the numerator and denominator by $a$ (which we know is not zero, so division is OK). $\endgroup$
    – Calum Gilhooley
    Commented Mar 5, 2015 at 22:20
  • $\begingroup$ See my answer for the R-rated version. $\endgroup$
    – marty cohen
    Commented Mar 5, 2015 at 22:43

2 Answers 2

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Explicitorizing Calum Gilhooley's answer (and possibly annoying some people):

$\begin{array}\\ r &=\frac{a+b}{2a+b}\\ &=\frac{2a+b-a}{2a+b}\\ &=1-\frac{a}{2a+b}\\ &=1-\frac{1}{2+b/a}\\ &=1-\frac{1}{2+x} \quad (x = b/a, 1/2 < x < 2)\\ \end{array} $

so $1-\frac14 > r > 1-\frac1{5/2} $ or $\frac35 < r < \frac34 $.

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Hint:

Express the fraction as a function of the single quantity $x = b/a$, which varies between $1/2$ and $2$. Then, rewrite this fractional expression $f(x)$ as an expression in which $x$ appears only once, so that you can more easily derive the maximum and minimum values of $f(x)$ from the known maximum and minimum values of $x$. These two bounds for $f(x)$ differ by $3/20$, as required.

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