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Question Number 56500 by kelly33 last updated on 17/Mar/19

find the X_0 value of the x variable with which the function assumes the lower value. f(x)=x^2 −3x+7+sin(πx)

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{X}}_{\mathrm{0}} \:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{x}}\:\boldsymbol{{variable}}\:\boldsymbol{{with}}\:\boldsymbol{{which}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{function}}\:\boldsymbol{{assumes}}\:\boldsymbol{{the}}\:\boldsymbol{{lower}}\:\boldsymbol{{value}}. \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7}+{sin}\left(\pi{x}\right) \\ $$

Commented by kelly33 last updated on 17/Mar/19

plz help me

$$ \\ $$$$ \\ $$$${plz}\:{help}\:{me} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

f(x)=x^2 −3x+7+sin(πx) =x^2 −2.x.(3/2)+(9/4)+7−(9/4)+sinπx =(x−(3/2))^2 +((19)/4)+sin(πx) for min value (x−(3/2))^2 =0 x=(3/2)→sin(((3π)/2))=−1 so min value is ((19)/4)−1=((15)/4)=3.75

$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7}+{sin}\left(\pi{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{7}−\frac{\mathrm{9}}{\mathrm{4}}+{sin}\pi{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{19}}{\mathrm{4}}+{sin}\left(\pi{x}\right) \\ $$$${for}\:{min}\:{value}\:\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\rightarrow{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)=−\mathrm{1} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{min}}\:\boldsymbol{{value}}\:\boldsymbol{{is}}\:\:\:\:\frac{\mathrm{19}}{\mathrm{4}}−\mathrm{1}=\frac{\mathrm{15}}{\mathrm{4}}=\mathrm{3}.\mathrm{75} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

Commented by kelly33 last updated on 19/Mar/19

=x^2 −2.x.(3/2)+(9/4)+7−(9/4)+sinπx =(x−(3/2))^2 +((19)/4)+sin(πx) where did this come from?

$$={x}^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{7}−\frac{\mathrm{9}}{\mathrm{4}}+{sin}\pi{x} \\ $$$$=\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{19}}{\mathrm{4}}+{sin}\left(\pi{x}\right) \\ $$$${where}\:{did}\:{this}\:{come}\:{from}? \\ $$$$ \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19

learn basic math then attemt hard problem.. formula (a+b)^2 =a^2 +2ab+b^2 now 4a^2 +4a+1 (2a)^2 +2.2a.1+(1)^2 (2a+1)^2 a^2 −3a+2 (a)^2 −2×a×(3/2)+((3/2))^2 −(9/4)+2 (a−(3/2))^2 −(1/4) similarly x^2 −3x+7 (x)^2 −2×x×(3/2)+((3/2))^2 +7−(9/4) (x−(3/2))^2 +((19)/4)

$${learn}\:{basic}\:{math}\:{then}\:{attemt}\:{hard}\:{problem}.. \\ $$$${formula}\:\:\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$${now}\: \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{1} \\ $$$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{2}{a}.\mathrm{1}+\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{3}{a}+\mathrm{2} \\ $$$$\left({a}\right)^{\mathrm{2}} −\mathrm{2}×{a}×\frac{\mathrm{3}}{\mathrm{2}}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{2} \\ $$$$\left({a}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${similarly} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7} \\ $$$$\left({x}\right)^{\mathrm{2}} −\mathrm{2}×{x}×\frac{\mathrm{3}}{\mathrm{2}}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7}−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{19}}{\mathrm{4}} \\ $$

Commented by kelly33 last updated on 19/Mar/19

where emerged (3/2) and ((−9)/4)

$$ \\ $$$${where}\:{emerged}\:\frac{\mathrm{3}}{\mathrm{2}}\:{and}\:\frac{−\mathrm{9}}{\mathrm{4}} \\ $$

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