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Question Number 92277 by  M±th+et+s last updated on 05/May/20

find a,b,c,d if f(x)=ax^3 +bx^2 +cx+d (3,3)is maximum value (5,1) is minimum value (4,2) is inflection point

$${find}\:{a},{b},{c},{d}\:\: \\ $$$${if}\:\:\:\:{f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$$\left(\mathrm{3},\mathrm{3}\right){is}\:{maximum}\:{value} \\ $$$$\left(\mathrm{5},\mathrm{1}\right)\:{is}\:{minimum}\:{value} \\ $$$$\left(\mathrm{4},\mathrm{2}\right)\:{is}\:{inflection}\:{point} \\ $$

Commented by  M±th+et+s last updated on 06/May/20

thank you sir but if we put x=3 f(3)=((27)/3)−4(9)+15(3)−((46)/3) f(3)=18−((46)/3)=(8/3) but in the question f(3)=3 ??

$${thank}\:{you}\:{sir}\:{but}\:{if}\:{we}\:{put}\:{x}=\mathrm{3}\: \\ $$$${f}\left(\mathrm{3}\right)=\frac{\mathrm{27}}{\mathrm{3}}−\mathrm{4}\left(\mathrm{9}\right)+\mathrm{15}\left(\mathrm{3}\right)−\frac{\mathrm{46}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{18}−\frac{\mathrm{46}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${but}\:{in}\:{the}\:{question}\:{f}\left(\mathrm{3}\right)=\mathrm{3}\:?? \\ $$$$\: \\ $$

Commented by john santu last updated on 06/May/20

it does mean question inconsistent f ′(x)= 3ax^2 +2bx+c = 0 has roots x=5 & x = 3 3a(x−5)(x−3)≡ 3ax^2 +2bx+c 3a(x^2 −8x+15)≡3ax^2 +2bx+c 2b=−24a ⇒ b = −12a c = 45a f(x) = ax^3 −12ax^2 +45a+d (3,3) ⇒ 3 =27a−108a+135a +d (4,2)⇒ 2= 64a−192a+90a+d

$$\mathrm{it}\:\mathrm{does}\:\mathrm{mean}\:\mathrm{question}\:\mathrm{inconsistent} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\:\mathrm{3ax}^{\mathrm{2}} +\mathrm{2bx}+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{roots}\:\mathrm{x}=\mathrm{5}\:\&\:\mathrm{x}\:=\:\mathrm{3}\: \\ $$$$\mathrm{3a}\left(\mathrm{x}−\mathrm{5}\right)\left(\mathrm{x}−\mathrm{3}\right)\equiv\:\mathrm{3ax}^{\mathrm{2}} +\mathrm{2bx}+\mathrm{c} \\ $$$$\mathrm{3a}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{15}\right)\equiv\mathrm{3ax}^{\mathrm{2}} +\mathrm{2bx}+\mathrm{c} \\ $$$$\mathrm{2b}=−\mathrm{24a}\:\Rightarrow\:\mathrm{b}\:=\:−\mathrm{12a} \\ $$$$\mathrm{c}\:=\:\mathrm{45a} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{ax}^{\mathrm{3}} −\mathrm{12ax}^{\mathrm{2}} +\mathrm{45a}+\mathrm{d} \\ $$$$\left(\mathrm{3},\mathrm{3}\right)\:\Rightarrow\:\mathrm{3}\:=\mathrm{27a}−\mathrm{108a}+\mathrm{135a}\:+\mathrm{d} \\ $$$$\left(\mathrm{4},\mathrm{2}\right)\Rightarrow\:\mathrm{2}=\:\mathrm{64a}−\mathrm{192a}+\mathrm{90a}+\mathrm{d} \\ $$

Commented by  M±th+et+s last updated on 06/May/20

i think the right solution is a=(1/2) b=−6 c=((45)/( 2)) d=−24 f(x)=(1/2)x^3 −6x^2 +((45)/2)x−24 f(3)=((27)/2)−54+((45)/2)(3)−24=3 f(5)=((125)/2)−150+((225)/2)−24=1 f(4)=32−96+90−24=2 f ′(x)=(3/2)x^2 −12x+((45)/2) f′(x)=0 x_1 =3 , x_2 =5 f′′(x)=3x−12 f′′(x)=0 x=4

$${i}\:{think}\:{the}\:{right}\:{solution}\:{is} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{b}=−\mathrm{6}\:\:{c}=\frac{\mathrm{45}}{\:\:\mathrm{2}}\:\:{d}=−\mathrm{24} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\frac{\mathrm{45}}{\mathrm{2}}{x}−\mathrm{24} \\ $$$${f}\left(\mathrm{3}\right)=\frac{\mathrm{27}}{\mathrm{2}}−\mathrm{54}+\frac{\mathrm{45}}{\mathrm{2}}\left(\mathrm{3}\right)−\mathrm{24}=\mathrm{3} \\ $$$${f}\left(\mathrm{5}\right)=\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{150}+\frac{\mathrm{225}}{\mathrm{2}}−\mathrm{24}=\mathrm{1} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{32}−\mathrm{96}+\mathrm{90}−\mathrm{24}=\mathrm{2} \\ $$$$ \\ $$$${f}\:'\left({x}\right)=\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{12}{x}+\frac{\mathrm{45}}{\mathrm{2}} \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\:\:\:{x}_{\mathrm{1}} =\mathrm{3}\:,\:{x}_{\mathrm{2}} =\mathrm{5} \\ $$$${f}''\left({x}\right)=\mathrm{3}{x}−\mathrm{12}\:\:\:\: \\ $$$${f}''\left({x}\right)=\mathrm{0}\:\:\:\:{x}=\mathrm{4} \\ $$

Commented by john santu last updated on 06/May/20

just two equations are enough. ����

Answered by MJS last updated on 06/May/20

ax^3 +bx^2 +cx+d=y 3ax^2 +2bx+c=y′ 6ax+2b=y′′ (3,3) ∈f(x) 27a+9b+3c+d=3 (I) (3,3) max 27a+6b+c=0 (II) (5,1) ∈f(x) 125a+25bb+5c+d=1 (III) (5,1) min 75a+10b+c=0 (IV) (4,2) ∈f(x) 64a+16b+4c+d=2 (V) (4,2) inflection 24a+2b=0 (VI) 6 equations and only 4 unknowns too much info might be inconsistent but in this case solving the system leads to a=(1/2), b=−6, c=((45)/2), d=−24

$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}={y} \\ $$$$\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}={y}' \\ $$$$\mathrm{6}{ax}+\mathrm{2}{b}={y}'' \\ $$$$\left(\mathrm{3},\mathrm{3}\right)\:\in{f}\left({x}\right) \\ $$$$\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=\mathrm{3}\:\left({I}\right) \\ $$$$\left(\mathrm{3},\mathrm{3}\right)\:\mathrm{max} \\ $$$$\mathrm{27}{a}+\mathrm{6}{b}+{c}=\mathrm{0}\:\left({II}\right) \\ $$$$\left(\mathrm{5},\mathrm{1}\right)\:\in{f}\left({x}\right) \\ $$$$\mathrm{125}{a}+\mathrm{25}{bb}+\mathrm{5}{c}+{d}=\mathrm{1}\:\left({III}\right) \\ $$$$\left(\mathrm{5},\mathrm{1}\right)\:\mathrm{min} \\ $$$$\mathrm{75}{a}+\mathrm{10}{b}+{c}=\mathrm{0}\:\left({IV}\right) \\ $$$$\left(\mathrm{4},\mathrm{2}\right)\:\in{f}\left({x}\right) \\ $$$$\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+{d}=\mathrm{2}\:\left({V}\right) \\ $$$$\left(\mathrm{4},\mathrm{2}\right)\:\mathrm{inflection} \\ $$$$\mathrm{24}{a}+\mathrm{2}{b}=\mathrm{0}\:\left({VI}\right) \\ $$$$\mathrm{6}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{only}\:\mathrm{4}\:\mathrm{unknowns} \\ $$$$\mathrm{too}\:\mathrm{much}\:\mathrm{info}\:\mathrm{might}\:\mathrm{be}\:\mathrm{inconsistent} \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{system}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}},\:{b}=−\mathrm{6},\:{c}=\frac{\mathrm{45}}{\mathrm{2}},\:{d}=−\mathrm{24} \\ $$

Commented by MJS last updated on 06/May/20

yes. we need only 4 equations to solve. if we have more we must check if they fit together

$$\mathrm{yes}.\:\mathrm{we}\:\mathrm{need}\:\mathrm{only}\:\mathrm{4}\:\mathrm{equations}\:\mathrm{to}\:\mathrm{solve}.\:\mathrm{if}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{more}\:\mathrm{we}\:\mathrm{must}\:\mathrm{check}\:\mathrm{if}\:\mathrm{they}\:\mathrm{fit}\:\mathrm{together} \\ $$

Commented by  M±th+et+s last updated on 06/May/20

thank you sir now it′s easy to solve from (1) and (3) we get −98a−16b−2c=2 .........(7) from (7) and (2) we get −44a−4b=2........(8) from (8) and (6) we get 2a=1⇒⇒a=(1/2)⇒⇒b=−6 from (2) we get c=((45)/2) from(1) we get d=−24

$${thank}\:{you}\:{sir}\:{now}\:{it}'{s}\:{easy}\:{to}\:{solve} \\ $$$${from}\:\:\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{3}\right)\:{we}\:{get}\: \\ $$$$−\mathrm{98}{a}−\mathrm{16}{b}−\mathrm{2}{c}=\mathrm{2}\:.........\left(\mathrm{7}\right) \\ $$$${from}\:\left(\mathrm{7}\right)\:{and}\:\left(\mathrm{2}\right)\:\:{we}\:{get}\: \\ $$$$−\mathrm{44}{a}−\mathrm{4}{b}=\mathrm{2}........\left(\mathrm{8}\right) \\ $$$${from}\:\left(\mathrm{8}\right)\:{and}\:\left(\mathrm{6}\right)\:{we}\:{get} \\ $$$$\mathrm{2}{a}=\mathrm{1}\Rightarrow\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\Rightarrow{b}=−\mathrm{6} \\ $$$${from}\:\left(\mathrm{2}\right)\:{we}\:{get}\:{c}=\frac{\mathrm{45}}{\mathrm{2}} \\ $$$${from}\left(\mathrm{1}\right)\:{we}\:{get}\:{d}=−\mathrm{24} \\ $$

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