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Question Number 122349 by liberty last updated on 16/Nov/20

Find the polynomial P(x) of least degree that has a maximum equal to 6 at x=1 and minimum equal to 2 at x=3.

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{of}\:\mathrm{least}\:\mathrm{degree} \\ $$$$\mathrm{that}\:\mathrm{has}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{6}\:\mathrm{at}\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{minimum}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2}\:\mathrm{at}\:\mathrm{x}=\mathrm{3}.\: \\ $$

Commented by benjo_mathlover last updated on 17/Nov/20

my answer : the polynomial of the least degree with roots x_1 =1 and x_2 =3 has the form a(x−1)(x−3). Hence P ′(x)=a(x−1)(x−3)=a(x^2 −4x+3) Since at the point x=1 there must be P(1)=6 , we have P(x)=∫_1 ^x P ′(x) dx + 6 P(x)= a∫_1 ^x (x^2 −4x+3)dx+6 P(x)= a((x^3 /3)−2x^2 +3x−(4/3))+6. The coefficient a is determined from the condition P(3)=2, whence a =3 , Hence P(x)=x^3 −6x^2 +9x+2.

$${my}\:{answer}\::\: \\ $$$${the}\:{polynomial}\:{of}\:{the}\:{least}\:{degree} \\ $$$${with}\:{roots}\:{x}_{\mathrm{1}} =\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\mathrm{3}\:{has}\:{the}\: \\ $$$${form}\:{a}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right).\:{Hence}\: \\ $$$${P}\:'\left({x}\right)={a}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)={a}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right) \\ $$$${Since}\:{at}\:{the}\:{point}\:{x}=\mathrm{1}\:{there}\:{must}\:{be} \\ $$$${P}\left(\mathrm{1}\right)=\mathrm{6}\:,\:{we}\:{have}\: \\ $$$${P}\left({x}\right)=\underset{\mathrm{1}} {\overset{{x}} {\int}}{P}\:'\left({x}\right)\:{dx}\:+\:\mathrm{6}\: \\ $$$${P}\left({x}\right)=\:{a}\underset{\mathrm{1}} {\overset{{x}} {\int}}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right){dx}+\mathrm{6}\: \\ $$$${P}\left({x}\right)=\:{a}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\frac{\mathrm{4}}{\mathrm{3}}\right)+\mathrm{6}. \\ $$$${The}\:{coefficient}\:{a}\:{is}\:{determined} \\ $$$${from}\:{the}\:{condition}\:{P}\left(\mathrm{3}\right)=\mathrm{2},\:{whence} \\ $$$${a}\:=\mathrm{3}\:,\:{Hence}\:{P}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{2}. \\ $$

Answered by mr W last updated on 17/Nov/20

f′(x)=k(x−1)(x−3)=k(x^2 −4x+3) f(x)=k((x^3 /3)−2x^2 +3x+c) f(1)=k((4/3)+c)=6 f(3)=kc=2 ⇒k=3 ⇒c=(2/3) ⇒f(x)=x^3 −6x^2 +9x+2

$${f}'\left({x}\right)={k}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)={k}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right) \\ $$$${f}\left({x}\right)={k}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+{c}\right) \\ $$$${f}\left(\mathrm{1}\right)={k}\left(\frac{\mathrm{4}}{\mathrm{3}}+{c}\right)=\mathrm{6} \\ $$$${f}\left(\mathrm{3}\right)={kc}=\mathrm{2} \\ $$$$\Rightarrow{k}=\mathrm{3} \\ $$$$\Rightarrow{c}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{2} \\ $$

Commented by mr W last updated on 16/Nov/20

Commented by liberty last updated on 16/Nov/20

yes...thanks...

$$\mathrm{yes}...\mathrm{thanks}... \\ $$

Commented by benjo_mathlover last updated on 17/Nov/20

typo sir. we get k=3 . it should be f(x)=3((x^3 /3)−2x^2 +3x+(2/3)) f(x) = x^3 −6x^3 +9x+2

$${typo}\:{sir}.\:{we}\:{get}\:{k}=\mathrm{3}\:.\:{it}\:{should} \\ $$$${be}\:{f}\left({x}\right)=\mathrm{3}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{9}{x}+\mathrm{2} \\ $$

Commented by mr W last updated on 17/Nov/20

yes, thanks!

$${yes},\:{thanks}! \\ $$

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