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

$$finda,b,c,d$$$$iff\left(x\right)={ax}^3+{bx}^2+cx+d$$$$(3,3)ismaximumvalue$$$$(5,1)isminimumvalue$$$$(4,2)isinflectionpoint$$

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

$$thankyousirbutifweputx=3$$$$f\left(3\right)=\frac{27}{3}-4\left(9\right)+15\left(3\right)-\frac{46}{3}$$$$f\left(3\right)=18-\frac{46}{3}=\frac{8}{3}$$$$butinthequestionf\left(3\right)=3??$$$$$$

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

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

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

$$ithinktherightsolutionis$$$$a=\frac{1}{2}b=-6c=\frac{45}{2}d=-24$$$$f\left(x\right)=\frac{1}{2}{x}^3-6x^2+\frac{45}{2}x-24$$$$f\left(3\right)=\frac{27}{2}-54+\frac{45}{2}\left(3\right)-24=3$$$$f\left(5\right)=\frac{125}{2}-150+\frac{225}{2}-24=1$$$$f\left(4\right)=32-96+90-24=2$$$$$$$$f{}^{\prime }\left(x\right)=\frac{3}{2}{x}^2-12x+\frac{45}{2}$$$$f^{\prime }\left(x\right)=0x_1=3,x_2=5$$$$f^{″}\left(x\right)=3x-12$$$$f^{″}\left(x\right)=0x=4$$

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

just two equations are enough. ����

Answered by MJS last updated on 06/May/20

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

Commented by MJS last updated on 06/May/20

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

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

$$thankyousirnow{it}^{\prime }seasytosolve$$$$from\left(1\right)and\left(3\right)weget$$$$-98a-16b-2c=2.........\left(7\right)$$$$from\left(7\right)and\left(2\right)weget$$$$-44a-4b=2........\left(8\right)$$$$from\left(8\right)and\left(6\right)weget$$$$2a=1\Rightarrow \Rightarrow a=\frac{1}{2}\Rightarrow \Rightarrow b=-6$$$$from\left(2\right)wegetc=\frac{45}{2}$$$$from\left(1\right)wegetd=-24$$

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