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Question Number 59114 by kelly33 last updated on 04/May/19
Commented by kelly33 last updated on 04/May/19
Please help me.
$${Please}\:{help}\:{me}. \\ $$
Answered by tanmay last updated on 04/May/19
5)trying p=4x+y^2 p=4x+4−2x^2 (dp/dx)=4−4x for max/min (dp/dx)=0→x=1 (d^2 p/dx^2 )=−4 (d^2 p/dx^2 )<0 so maximum y^2 =4−2x^2 →y^2 =4−2(1)^2 y=±(√2) p=4x+y^2 =4(1)+2=6 (4x+y^2 )_(max) =6 another way p=4x+4−2x^2 p=−2(x^2 −2x+1)+2+4 p=−2(x−1)^2 +6 p=6−2(x−1)^2 since (x−1)^2 ≥0 so p_(max) =6 p_(min) =not feasible
$$\left.\mathrm{5}\right){trying}\:{p}=\mathrm{4}{x}+{y}^{\mathrm{2}} \\ $$$${p}=\mathrm{4}{x}+\mathrm{4}−\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\frac{{dp}}{{dx}}=\mathrm{4}−\mathrm{4}{x}\:{for}\:{max}/{min}\:\frac{{dp}}{{dx}}=\mathrm{0}\rightarrow{x}=\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{2}} {p}}{{dx}^{\mathrm{2}} }=−\mathrm{4}\:\:\:\:\frac{{d}^{\mathrm{2}} {p}}{{dx}^{\mathrm{2}} }<\mathrm{0}\:\:{so}\:{maximum} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}{x}^{\mathrm{2}} \rightarrow{y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$${y}=\pm\sqrt{\mathrm{2}}\: \\ $$$${p}=\mathrm{4}{x}+{y}^{\mathrm{2}} \\ $$$$=\mathrm{4}\left(\mathrm{1}\right)+\mathrm{2}=\mathrm{6} \\ $$$$\left(\mathrm{4}{x}+{y}^{\mathrm{2}} \right)_{{max}} =\mathrm{6} \\ $$$${another}\:{way} \\ $$$${p}=\mathrm{4}{x}+\mathrm{4}−\mathrm{2}{x}^{\mathrm{2}} \\ $$$${p}=−\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{2}+\mathrm{4} \\ $$$${p}=−\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6} \\ $$$${p}=\mathrm{6}−\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${since}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${so}\:{p}_{{max}} =\mathrm{6} \\ $$$${p}_{{min}} ={not}\:{feasible} \\ $$$$ \\ $$
Answered by tanmay last updated on 04/May/19
4)f(x)=∣x−1∣+∣x^2 −2x∣ f(x)=∣x−1∣+∣x(x−2)∣ critical value of x are 0,1 snd 2 f(x)=1 when x=2 f(x)=(x−1)−x(x−2) when 2>x>1 =−x^2 +2x+x−1 =−x^2 +3x−1 =1 when x=1 = −(x−1)−x(x−2) when 1>x>0 =−x+1−x^2 +2x =−x^2 +3x+1 f(x)=1 when x=0 now f(x)=−x^2 +3x−1 when 2>x>1 (df/dx)=−2x+3 for max/min (df/dx)=0 so x=(3/2)=1.5 (d^2 f/dx^2 )=−2<0 so at x=(3/2) f(x) maximum f(x)_(max) =−x^2 +3x−1 =−((3/2))^2 +3×(3/2)−1 =((−9)/4)+(9/2)−1 =((−9+18−4)/4)=(5/4) f(x)=−x^2 +3x+1 when 1>x>0 (df/dx)=−2x+3 for max/min (df/dx)=0 so x=(3/2) (d^2 f/dx^2 )=−2<0 maximum at x=(3/2) but x=(3/2) does not lie in the interval [that is 1>x>0 conclusion f(x)_(max) =(5/4) at x=(3/2) f(x)_(min) =1 when x =2 f(x)_(min) =1 when x=1 f(x)_(min) =1 when x=0 f(x)_(max) =(5/4) when x=(3/2)
$$\left.\mathrm{4}\right){f}\left({x}\right)=\mid{x}−\mathrm{1}\mid+\mid{x}^{\mathrm{2}} −\mathrm{2}{x}\mid \\ $$$${f}\left({x}\right)=\mid{x}−\mathrm{1}\mid+\mid{x}\left({x}−\mathrm{2}\right)\mid \\ $$$${critical}\:{value}\:{of}\:{x}\:{are}\:\mathrm{0},\mathrm{1}\:{snd}\:\mathrm{2} \\ $$$$ \\ $$$${f}\left({x}\right)=\mathrm{1}\:\:{when}\:{x}=\mathrm{2} \\ $$$$ \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)−{x}\left({x}−\mathrm{2}\right)\:\:{when}\:\mathrm{2}>{x}>\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:=−{x}^{\mathrm{2}} +\mathrm{2}{x}+{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}\:\:\:{when}\:{x}=\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\:−\left({x}−\mathrm{1}\right)−{x}\left({x}−\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:{when}\:\mathrm{1}>{x}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−{x}+\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\:\:\:{f}\left({x}\right)=\mathrm{1}\:\:\:\:{when}\:{x}=\mathrm{0} \\ $$$$ \\ $$$${now}\:{f}\left({x}\right)=−\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}−\mathrm{1}\:\:\:\boldsymbol{{when}}\:\mathrm{2}>\boldsymbol{{x}}>\mathrm{1} \\ $$$$\frac{\boldsymbol{{df}}}{\boldsymbol{{dx}}}=−\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\:\:\:{for}\:{max}/{min}\:\frac{{df}}{{dx}}=\mathrm{0}\:{so}\:{x}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5} \\ $$$$\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{f}}}{\boldsymbol{{dx}}^{\mathrm{2}} }=−\mathrm{2}<\mathrm{0} \\ $$$${so}\:{at}\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\:{f}\left({x}\right)\:{maximum} \\ $$$${f}\left({x}\right)_{{max}} =−{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}×\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{9}+\mathrm{18}−\mathrm{4}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\:{when}\:\:\:\mathrm{1}>{x}>\mathrm{0} \\ $$$$\frac{{df}}{{dx}}=−\mathrm{2}{x}+\mathrm{3}\:\:{for}\:{max}/{min}\:\frac{{df}}{{dx}}=\mathrm{0}\:{so}\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=−\mathrm{2}<\mathrm{0}\:\:{maximum}\:{at}\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\:{but}\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\:{does}\:{not} \\ $$$${lie}\:\:{in}\:{the}\:{interval}\:\left[{that}\:{is}\:\:\mathrm{1}>{x}>\mathrm{0}\right. \\ $$$$\:\boldsymbol{{conclusion}}\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{max}}} =\frac{\mathrm{5}}{\mathrm{4}}\:\:\boldsymbol{{at}}\:\boldsymbol{{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{min}}} =\mathrm{1}\:\boldsymbol{{when}}\:\boldsymbol{{x}}\:=\mathrm{2}\: \\ $$$$\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{min}}} =\mathrm{1}\:\:\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{1} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{min}}} =\mathrm{1}\:\:\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{max}}} =\frac{\mathrm{5}}{\mathrm{4}}\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by tanmay last updated on 05/May/19
3)y=ax^3 +bx^2 +cx+d (0,1) (−1,−2),(1,2)(2,9) 1=d −2=−a+b−c+1 −a+b−c=−3 2=a+b+c+1 −a+b−c=−3 a+b+c=1 2b=−2 b=−1 so −a−1−c=−3 a+c=2.... a−1+c=1 a+c=2 9=a(2)^3 +b(2)^2 +c(2)+d 9=8a−4+2c+1 8a+2c=12 4a+c=6 a+c=2 3a=4 a=(4/3) c=2−(4/3)→ c=(2/3) y=(4/3)x^3 −x^2 +(2/3)x+1
$$\left.\mathrm{3}\right){y}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$$\left(\mathrm{0},\mathrm{1}\right)\:\left(−\mathrm{1},−\mathrm{2}\right),\left(\mathrm{1},\mathrm{2}\right)\left(\mathrm{2},\mathrm{9}\right) \\ $$$$\mathrm{1}={d} \\ $$$$−\mathrm{2}=−{a}+{b}−{c}+\mathrm{1} \\ $$$$−{a}+{b}−{c}=−\mathrm{3} \\ $$$$\mathrm{2}={a}+{b}+{c}+\mathrm{1} \\ $$$$−{a}+{b}−{c}=−\mathrm{3} \\ $$$$\:\:\:\:{a}+{b}+{c}=\mathrm{1} \\ $$$$\mathrm{2}{b}=−\mathrm{2}\:\:\:{b}=−\mathrm{1} \\ $$$${so}\: \\ $$$$−{a}−\mathrm{1}−{c}=−\mathrm{3} \\ $$$${a}+{c}=\mathrm{2}…. \\ $$$$ \\ $$$${a}−\mathrm{1}+{c}=\mathrm{1} \\ $$$${a}+{c}=\mathrm{2} \\ $$$$\mathrm{9}={a}\left(\mathrm{2}\right)^{\mathrm{3}} +{b}\left(\mathrm{2}\right)^{\mathrm{2}} +{c}\left(\mathrm{2}\right)+{d} \\ $$$$\mathrm{9}=\mathrm{8}{a}−\mathrm{4}+\mathrm{2}{c}+\mathrm{1} \\ $$$$\mathrm{8}{a}+\mathrm{2}{c}=\mathrm{12} \\ $$$$\mathrm{4}{a}+{c}=\mathrm{6} \\ $$$${a}+{c}=\mathrm{2} \\ $$$$\mathrm{3}{a}=\mathrm{4} \\ $$$${a}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${c}=\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}}\rightarrow\:\:\:{c}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${y}=\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{x}+\mathrm{1} \\ $$
Answered by tanmay last updated on 05/May/19
2)formula X=(x−h)cosθ+(y−k)sinθ Y=−(x−h)sinθ+(y−k)cosθ here (h,k)→(2,1) θ=45^(o ) (x,y)=(3,3) X=(3−2)cos45^o +(3−1)sin45^o X=(1/( (√2)))(1+2)=(3/( (√2))) Y=−(3−2)sin45^o +(3−1)cos45^o Y=(1/( (√2)))(−1+2)=(1/( (√2)))
$$\left.\mathrm{2}\right){formula} \\ $$$${X}=\left({x}−{h}\right){cos}\theta+\left({y}−{k}\right){sin}\theta \\ $$$$\:{Y}=−\left({x}−{h}\right){sin}\theta+\left({y}−{k}\right){cos}\theta \\ $$$${here}\:\left({h},{k}\right)\rightarrow\left(\mathrm{2},\mathrm{1}\right)\:\:\theta=\mathrm{45}^{{o}\:\:} \\ $$$$\left({x},{y}\right)=\left(\mathrm{3},\mathrm{3}\right) \\ $$$${X}=\left(\mathrm{3}−\mathrm{2}\right){cos}\mathrm{45}^{{o}} +\left(\mathrm{3}−\mathrm{1}\right){sin}\mathrm{45}^{{o}} \\ $$$${X}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+\mathrm{2}\right)=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$$${Y}=−\left(\mathrm{3}−\mathrm{2}\right){sin}\mathrm{45}^{{o}} +\left(\mathrm{3}−\mathrm{1}\right){cos}\mathrm{45}^{{o}} \\ $$$${Y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(−\mathrm{1}+\mathrm{2}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\: \\ $$

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