Menu Close

find-minimum-value-of-function-f-x-x-17-3-x-x-gt-0-




Question Number 114753 by bemath last updated on 21/Sep/20
find minimum value of function  f(x) = (((x+17)^3 )/x) , x>0
$${find}\:{minimum}\:{value}\:{of}\:{function} \\ $$$${f}\left({x}\right)\:=\:\frac{\left({x}+\mathrm{17}\right)^{\mathrm{3}} }{{x}}\:,\:{x}>\mathrm{0} \\ $$
Answered by Olaf last updated on 21/Sep/20
f′(x) = ((3(x+17)^2 x−(x+17)^3 )/x^2 )  f′(x) = (((x+17)^2 (3x−(x+17)))/x^2 )  f′(x) = (((x+17)^2 (2x−17))/x^2 )  f′(x) = 0 ⇔ x = ((17)/2) (x > 0)  If x < ((17)/2), f′(x) < 0  If x > ((17)/2), f′(x) > 0  ⇒ x = ((17)/2) is a minimum
$${f}'\left({x}\right)\:=\:\frac{\mathrm{3}\left({x}+\mathrm{17}\right)^{\mathrm{2}} {x}−\left({x}+\mathrm{17}\right)^{\mathrm{3}} }{{x}^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)\:=\:\frac{\left({x}+\mathrm{17}\right)^{\mathrm{2}} \left(\mathrm{3}{x}−\left({x}+\mathrm{17}\right)\right)}{{x}^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)\:=\:\frac{\left({x}+\mathrm{17}\right)^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{17}\right)}{{x}^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:\frac{\mathrm{17}}{\mathrm{2}}\:\left({x}\:>\:\mathrm{0}\right) \\ $$$$\mathrm{If}\:{x}\:<\:\frac{\mathrm{17}}{\mathrm{2}},\:{f}'\left({x}\right)\:<\:\mathrm{0} \\ $$$$\mathrm{If}\:{x}\:>\:\frac{\mathrm{17}}{\mathrm{2}},\:{f}'\left({x}\right)\:>\:\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:\frac{\mathrm{17}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{minimum} \\ $$$$ \\ $$
Commented by bemath last updated on 21/Sep/20
give thanks
$${give}\:{thanks}\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *