Question Number 62211 by Tony Lin last updated on 18/Jun/19
$$\frac{{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+\mathrm{3}}{Max}=\frac{\mathrm{5}}{\mathrm{3}}? \\ $$
Commented by MJS last updated on 17/Jun/19
$$\mathrm{what}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}? \\ $$
Commented by Tony Lin last updated on 17/Jun/19
$${how}\:{to}\:{find}\:{the}\:{maximum}\:{value}\:{of}\: \\ $$$$\frac{{x}}{\sqrt{\mathrm{4}+{x}^{\mathrm{2}} }−\mathrm{3}}\:? \\ $$
Commented by MJS last updated on 18/Jun/19
![f(x)=(x/((√(4−x^2 ))+3)) is defined for x∈[−2; 2] the derivate is not defined for the borders ⇒ if there′s no other minimum or maximum look at these borders f(2)=(2/3) is the maximum because f′(x)=((4+3(√(4−x^2 )))/((3+(√(4−x^2 )))(√(4−x^2 ))))>0 ∀x∈]−2; 2[](Q62231.png)
$${f}\left({x}\right)=\frac{{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+\mathrm{3}}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\left[−\mathrm{2};\:\mathrm{2}\right] \\ $$$$\mathrm{the}\:\mathrm{derivate}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:\mathrm{the}\:\mathrm{borders} \\ $$$$\Rightarrow\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{other}\:\mathrm{minimum}\:\mathrm{or}\:\mathrm{maximum} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{these}\:\mathrm{borders} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{because} \\ $$$$\left.{f}'\left({x}\right)=\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\left(\mathrm{3}+\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}>\mathrm{0}\:\forall{x}\in\right]−\mathrm{2};\:\mathrm{2}\left[\right. \\ $$