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Question Number 39607 by Rio Mike last updated on 08/Jul/18
find the minimum and maximum value of the quadratic functions a) 4x^2 + 5x + 1 b) x + (2/x) = 3 c) x^2 − (x/4) + 6 hence draw each draw
$${find}\:{the}\: \\ $$$${minimum}\:{and}\:{maximum}\:{value} \\ $$$${of}\:{the}\:{quadratic}\:{functions} \\ $$$$\left.{a}\right)\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\:\mathrm{1} \\ $$$$\left.{b}\right)\:{x}\:+\:\frac{\mathrm{2}}{{x}}\:=\:\mathrm{3} \\ $$$$\left.{c}\right)\:{x}^{\mathrm{2}} \:−\:\frac{{x}}{\mathrm{4}}\:+\:\mathrm{6} \\ $$$${hence}\:{draw}\:{each}\:{draw} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18
a)4x^2 +5x+1 4(x^2 +(5/4)x+(1/4)) 4{(x^2 +2.x.(5/8)+((25)/(64))+(1/4)−((25)/(64)))} 4{(x+(5/8))^2 +((16−25)/(64))} 4{(x+(5/8))^2 +((−9)/(64))} 4(x+(5/8))^2 −(9/(16)) (x+(5/8))^2 >0 as the value of x increases so the value of (x+(5/8))^2 increases...hence 4x^2 +5x+1 has no maximum value..its minimum value is ((−9)/(16)) when x=((−5)/8) other method... by using calculus y=4x^2 +5x+1 (dy/dx)=8x+5 for min/max (dy/dx)=0 so x=−(5/8) (d^2 y/dx^2 )=8 that means (d^2 y/dx^2 )>0 so 4x^2 +5x+1 has minimum value at x=−(5/8) 4.(((25)/(64)))+5(((−5)/8))+1 ((25)/(16))−((25)/8)+1 ((25−50+16)/(16))=((−9)/(16))
$$\left.{a}\right)\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{4}\left\{\left({x}^{\mathrm{2}} +\mathrm{2}.{x}.\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{25}}{\mathrm{64}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{25}}{\mathrm{64}}\right)\right\} \\ $$$$\mathrm{4}\left\{\left({x}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} +\frac{\mathrm{16}−\mathrm{25}}{\mathrm{64}}\right\} \\ $$$$\mathrm{4}\left\{\left({x}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} +\frac{−\mathrm{9}}{\mathrm{64}}\right\} \\ $$$$\mathrm{4}\left({x}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$\left({x}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$${as}\:{the}\:{value}\:{of}\:{x}\:{increases}\:{so}\:{the}\:{value}\:{of}\: \\ $$$$\left({x}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} \:{increases}…{hence}\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\:{has}\:{no} \\ $$$${maximum}\:{value}..{its}\:{minimum}\:{value}\:{is} \\ $$$$\frac{−\mathrm{9}}{\mathrm{16}}\:{when}\:{x}=\frac{−\mathrm{5}}{\mathrm{8}} \\ $$$$\boldsymbol{{other}}\:\boldsymbol{{method}}… \\ $$$${by}\:{using}\:{calculus} \\ $$$${y}=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{8}{x}+\mathrm{5} \\ $$$${for}\:{min}/{max}\:\frac{{dy}}{{dx}}=\mathrm{0}\:\:\:{so}\:{x}=−\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{8}\:\:\:{that}\:{means}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }>\mathrm{0} \\ $$$${so}\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\:\:{has}\:{minimum}\:{value}\:{at}\:{x}=−\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$\mathrm{4}.\left(\frac{\mathrm{25}}{\mathrm{64}}\right)+\mathrm{5}\left(\frac{−\mathrm{5}}{\mathrm{8}}\right)+\mathrm{1} \\ $$$$\frac{\mathrm{25}}{\mathrm{16}}−\frac{\mathrm{25}}{\mathrm{8}}+\mathrm{1}\:\: \\ $$$$\frac{\mathrm{25}−\mathrm{50}+\mathrm{16}}{\mathrm{16}}=\frac{−\mathrm{9}}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18
c)y=((4x^2 −x+24)/4) (1/4){(2x)^2 −2.2x.(1/4)+(1/(16))+24−(1/(16))} (1/4){(2x−(1/4))^2 +((24×16−1)/(16))} as the value of x increases so the value of expression increases..so the expression has no maximum value.. its minimum value is ((24×16−1)/(4×16)) when x=(1/8) min value=((383)/(64)) when x=(1/8) by calculus y=x^2 −(x/4)+6 (dy/dx)=2x−(1/4) for min/max (dy/dx)=0 so x=(1/8) (d^2 y/dx^2 )=2 so (d^2 y/dx^2 )>0 so the expression has min value at x=(1/8) min valud ((1/8))^2 −(1/(32))+6 ((1−2+384)/(64)) =((383)/(64))
$$\left.{c}\right){y}=\frac{\mathrm{4}{x}^{\mathrm{2}} −{x}+\mathrm{24}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left\{\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{2}.\mathrm{2}{x}.\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+\mathrm{24}−\frac{\mathrm{1}}{\mathrm{16}}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left\{\left(\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{24}×\mathrm{16}−\mathrm{1}}{\mathrm{16}}\right\} \\ $$$${as}\:{the}\:{value}\:{of}\:{x}\:{increases}\:{so}\:{the}\:{value}\:{of} \\ $$$${expression}\:{increases}..{so}\:{the}\:{expression}\:{has} \\ $$$${no}\:{maximum}\:{value}.. \\ $$$${its}\:{minimum}\:{value}\:{is}\:\frac{\mathrm{24}×\mathrm{16}−\mathrm{1}}{\mathrm{4}×\mathrm{16}}\:\:{when}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${min}\:{value}=\frac{\mathrm{383}}{\mathrm{64}}\:\:{when}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{calculus}} \\ $$$${y}={x}^{\mathrm{2}} −\frac{{x}}{\mathrm{4}}+\mathrm{6} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${for}\:{min}/{max}\:\frac{{dy}}{{dx}}=\mathrm{0}\:\:\:{so}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }=\mathrm{2}\:\:\:\:\:\:\:\:{so}\:\:\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{{dx}^{\mathrm{2}} }>\mathrm{0} \\ $$$${so}\:{the}\:{expression}\:{has}\:{min}\:{value}\:{at}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${min}\:{valud}\:\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{32}}+\mathrm{6} \\ $$$$\frac{\mathrm{1}−\mathrm{2}+\mathrm{384}}{\mathrm{64}}\:=\frac{\mathrm{383}}{\mathrm{64}} \\ $$
Answered by MJS last updated on 08/Jul/18
y=ax^2 +bx+c zeros at x=−(b/(2a))±((√(b^2 −4ac))/(2a)) a>0 ⇒ min at x=−(b/(2a)); y=((−b^2 +4ac)/(4a^2 )) a<0 ⇒ max at x=−(b/(2a)); y=((−b^2 +4ac)/(4a^2 )) y=x^2 +px+q ⇒ min at x=−(p/2); y=((−p^2 +4q)/4) zeros at x=−(p/2)±((√(p^2 −4q))/2) y=−x^2 +px+q ⇒ max at x=(p/2); y=((p^2 +4q)/4) zeros at x=(p/2)±((√(p^2 +4q))/2)
$${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${zeros}\:{at}\:{x}=−\frac{{b}}{\mathrm{2}{a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${a}>\mathrm{0}\:\Rightarrow\:{min}\:{at}\:{x}=−\frac{{b}}{\mathrm{2}{a}};\:{y}=\frac{−{b}^{\mathrm{2}} +\mathrm{4}{ac}}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${a}<\mathrm{0}\:\Rightarrow\:{max}\:{at}\:{x}=−\frac{{b}}{\mathrm{2}{a}};\:{y}=\frac{−{b}^{\mathrm{2}} +\mathrm{4}{ac}}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${y}={x}^{\mathrm{2}} +{px}+{q}\:\Rightarrow\:{min}\:{at}\:{x}=−\frac{{p}}{\mathrm{2}};\:{y}=\frac{−{p}^{\mathrm{2}} +\mathrm{4}{q}}{\mathrm{4}} \\ $$$${zeros}\:{at}\:{x}=−\frac{{p}}{\mathrm{2}}\pm\frac{\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}} \\ $$$${y}=−{x}^{\mathrm{2}} +{px}+{q}\:\Rightarrow\:{max}\:{at}\:{x}=\frac{{p}}{\mathrm{2}};\:{y}=\frac{{p}^{\mathrm{2}} +\mathrm{4}{q}}{\mathrm{4}} \\ $$$${zeros}\:{at}\:{x}=\frac{{p}}{\mathrm{2}}\pm\frac{\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}} \\ $$

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