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Question Number 114753 by bemath last updated on 21/Sep/20

$$findminimumvalueoffunction$$ $$f\left(x\right)=\frac{{(x+17)}^3}{x},x>0$$

Answered by Olaf last updated on 21/Sep/20

$$f^{\prime }\left(x\right)=\frac{3{(x+17)}^{2}x-{(x+17)}^3}{x^2}$$ $$f^{\prime }\left(x\right)=\frac{{(x+17)}^2(3x-(x+17))}{x^2}$$ $$f^{\prime }\left(x\right)=\frac{{(x+17)}^2(2x-17)}{x^2}$$ $$f^{\prime }\left(x\right)=0\iff x=\frac{17}{2}(x>0)$$ $$\mathrm{If}x<\frac{17}{2},f^{\prime }\left(x\right)<0$$ $$\mathrm{If}x>\frac{17}{2},f^{\prime }\left(x\right)>0$$ $$\Rightarrow x=\frac{17}{2}\mathrm{is}\mathrm{a}\mathrm{minimum}$$ $$$$

Commented bybemath last updated on 21/Sep/20

$$givethanks$$

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