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Options : $b^2-4ac > 0$

$b^2/4ac >0$

$b/(c-1) >0$

$c/(a-1)>0$

$a/(b-1)>0$

Relative max or min is local max or min, means max or min at certain open interval..

$xy + ax^2 + bx + c = 0 $

$ax^2 + (y+b)x + c = 0 $ if its a qudratic fuction. Then the max or min point will be (-b/2a , f(-b/2a))

$y = (-ax^2 - bx - c)/x$ But what does it mean by $x+y$ ? When $x+y$ will have relative max or min?

Dini
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1 Answers1

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Get $y$ by itself so that $$y = \frac{-ax^{2}-bx-c}{x}$$ Then substitute this for $y$ in $f(x) = x+y$ so we have $$f(x) = \frac{-ax^{2}-bx-c}{x} + x$$ and we want to know if $f(x)$ will have a relative minimum or maximum. The cases where it will not have a minimum or maximum are when the derivative has no real zeroes. Lets find the derivative using the product rule: $$f'(x) = \frac{1}{x}(-2ax-b)-\frac{1}{x^{2}}(-ax^{2}-bx-c)+1$$ Simplify fractions $$f'(x) = \frac{x(-2ax-b)-(-ax^{2}-bx-c)+x^{2}}{x^{2}}$$ Simplify $$f'(x) = \frac{(-a+1)x^{2} + c}{x^{2}}$$ We wish to know when the derivative will be $0$, and we can multiply away the denominator to get $$0 = (-a+1)x^{2} + c$$ $$x^{2} = \frac{-c}{-a+1} = \frac{c}{a-1}$$ Notice that we need real roots in order to have a relative minimum or maximum, and we are taking a square root, thus $$\frac{c}{a-1}>0$$

Ty Jensen
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