Let It Change

Algebra Level 3

If in the formula A = e q S + q r \displaystyle A= \frac { eq }{ S+qr } , q is increased while e , S and r are kept constant, then A:

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increases then decreases increases decreases and then increases decreases

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1 solution

Soumo Mukherjee
Nov 2, 2014

A = e q S + q r = e S / q + r A=\frac { eq }{ S+qr } =\frac { e }{ S/q+r }

Therefore an increase in q makes the

denominator of the fraction smaller and, hence, the fraction A becomes larger.

Hey, Math Philic!

You think what you did back at Zombie Logs was stupid? LOOK AT WHAT I DID!!


Oh the lack of Calculus in this makes analysis so painful!

Nice problem, by the way!

But I gotta tell you: I got myself so confused on this! So, an important distinction: While the function always increases in value, it's magnitude doesn't.

So a little note: While it may be tempting to say that for all q < 1 \left| q \right| < 1 , the denominator actually becomes larger (which is what I did). But notice that talking about the behavior of 1 / q 1/q alone is meaningless if we don't account for r r , because as far as A A is concerned, A A becomes larger/smaller only when the entire denominator becomes smaller/larger. A A will always display an infinity (asymptotic) behavior around S / q + r = 0 S/q+r=0 , and always slowly level off afterwards (which makes sense, because S / q S/q becomes smaller and smaller, making S / q S/q smaller and smaller too, much at a much slower rate than around the asymptote (for which the denominator takes absolute values < 0 <0 (our (or my) original confusion resolved).

Take a look at some asymptotic behaviors for different graphs:

s s

ss ss

(the tiny function to the right is the first graph, btw)

John M. - 6 years, 7 months ago

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Those things never occurred to me!

"careless thinking is always dangerous thing "

I need your help..your help..your help! here I myself don't know accurate answers :(. Nobody answered them even :'(

Soumo Mukherjee - 6 years, 7 months ago

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Eh I think you should rephrase that to "careless thinking" or "pointless scrutiny". Little knowledge implies I don't know much about basic functions likes this (not true!) :/

Alrigh leemm e take a lok

John M. - 6 years, 7 months ago

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