Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 157116 by cortano last updated on 20/Oct/21

 max ∧ min of f(x) =(√x) +4(√((1−x)/2))

$$\:{max}\:\wedge\:{min}\:{of}\:{f}\left({x}\right)\:=\sqrt{{x}}\:+\mathrm{4}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{2}}} \\ $$

Commented by cortano last updated on 20/Oct/21

without derivative

$${without}\:{derivative} \\ $$

Commented by mr W last updated on 20/Oct/21

max=3  min=1

$${max}=\mathrm{3} \\ $$$${min}=\mathrm{1} \\ $$

Answered by mr W last updated on 20/Oct/21

we see 0≤x≤1, f(x)≥0  so let x=sin^2  θ with 0≤θ≤(π/2)  f(x)=sin θ+2(√2)cos θ=3 sin (θ+tan^(−1) 2(√2))  max=3 at θ=(π/2)−tan^(−1) 2(√2) or x=cos^2  (tan^(−1) 2(√2))=(1/9)  min=min{f(0),f(1)}=f(1)=1

$${we}\:{see}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1},\:{f}\left({x}\right)\geqslant\mathrm{0} \\ $$$${so}\:{let}\:{x}=\mathrm{sin}^{\mathrm{2}} \:\theta\:{with}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\mathrm{sin}\:\theta+\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}\:\theta=\mathrm{3}\:\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${max}=\mathrm{3}\:{at}\:\theta=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\mathrm{2}}\:{or}\:{x}=\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${min}={min}\left\{{f}\left(\mathrm{0}\right),{f}\left(\mathrm{1}\right)\right\}={f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com