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Find the minimum value and the maximum value of the function $$y(x)=\frac{x^3}{x-3}$$ when $4\le x\le5$

I found that $f(x)$ is decreasing on the interval $[4,\frac{9}{2}]$ and increasing on $[\frac{9}{2},5]$ What i actually want is, find way to solve problem by original way, with something like simple formulas and deduction, Could someone show it?

when $4\le x\le5$

I tried to give numerator +27-27 and cut fraction and try to find min from quadratic equation, but i cannt achive something, probably its bad way

Américo Tavares
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Музаффар Шакаров
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1 Answers1

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To find the minimum we can do the following: $$\frac{x^3}{x-3}-m=\frac{x^3-mx+3m}{x-3}$$ and now our goal is to find $m$ such that the numerator has a double root. That is, $$x^3-mx+3m=(x-a)^2(x-b)$$

This yields the system: $$b=-2a$$ $$a^2+2ab=-m$$ $$a^2b=-3m$$ We find that $a=\frac92$, $b=-9$ and $m=\frac{243}4$.

ajotatxe
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