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Question Number 202795 by mathlove last updated on 03/Jan/24

if f(−1)=f(0)=f(2)=0 and f(1)=6 then find f(x)=?

$${if}\:{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{6} \\ $$$${then}\:{find}\:{f}\left({x}\right)=? \\ $$

Answered by mr W last updated on 03/Jan/24

f(−1)=f(0)=f(2)=0: ⇒f(x)=x(x+1)(x−2)p(x) f(1)=6: ⇒f(1)=1×(1+1)(1−2)p(1)=6 ⇒p(1)=−3 ⇒p(x)=(x−4)q(x) with q(1)=1 solution: f(x)=x(x+1)(x−2)(x−4)q(x) with q(x)=any function and q(1)=1.

$${f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)=\mathrm{0}: \\ $$$$\Rightarrow{f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right){p}\left({x}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{6}: \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1}×\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}−\mathrm{2}\right){p}\left(\mathrm{1}\right)=\mathrm{6} \\ $$$$\Rightarrow{p}\left(\mathrm{1}\right)=−\mathrm{3} \\ $$$$\Rightarrow{p}\left({x}\right)=\left({x}−\mathrm{4}\right){q}\left({x}\right)\:{with}\:{q}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${solution}: \\ $$$${f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right){q}\left({x}\right) \\ $$$${with}\:{q}\left({x}\right)={any}\:{function}\:{and}\:{q}\left(\mathrm{1}\right)=\mathrm{1}. \\ $$

Commented by mathlove last updated on 03/Jan/24

thanks mr W

$${thanks}\:{mr}\:{W} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Jan/24

f(x)=x(x+1)(x−2)(x−4)^(?) q(x) Thanx in advance sir!

$${f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\overset{?} {\left({x}−\mathrm{4}\right)}{q}\left({x}\right) \\ $$$$\mathcal{T}{hanx}\:{in}\:{advance}\:{sir}! \\ $$

Commented by mr W last updated on 03/Jan/24

such that f(1)=6

$${such}\:{that}\:{f}\left(\mathrm{1}\right)=\mathrm{6} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Jan/24

ThanX again sir!

$$\mathcal{T}{han}\mathcal{X}\:{again}\:\boldsymbol{{sir}}! \\ $$

Answered by a.lgnaoui last updated on 03/Jan/24

−1;0 et 2 racines de f(x) ⇒f(x)=ax(x+1)(x−2) f(1)=a×2×(−1)=−2a=6 ⇒ a=−(6/2)=−3 alors f(x)=−3x(x+1)(x−2) f(x)=−3x(x^2 −x−2)

$$\:−\mathrm{1};\mathrm{0}\:\mathrm{et}\:\mathrm{2}\:\mathrm{racines}\:\mathrm{de}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right) \\ $$$$\:\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{a}×\mathrm{2}×\left(−\mathrm{1}\right)=−\mathrm{2a}=\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{a}=−\frac{\mathrm{6}}{\mathrm{2}}=−\mathrm{3} \\ $$$$\mathrm{alors}\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=−\mathrm{3}\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right) \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=−\mathrm{3}\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}−\mathrm{2}\right) \\ $$

Answered by Frix last updated on 04/Jan/24

Usually we′re looking for the polynome of minimal degree. With n equations we get a polynome of degree n−1 ⇒ with 4 equations we get a polynome of degree 3. The usual method is this: ax^3 +bx^2 +cx+d=f(x) f(0)=0 ⇒ d=0 (1) f(−1)=0 −a+b−c=0 (2) f(2)=0 8a+4b+2c=0 (3) f(1)=6 a+b+c=6 (4) (2) ⇒ b=a+c (3) 12a+6c=0 ⇒ c=−2a ⇒ b=−a (4) −2a=6 ⇒ a=−3 ⇒ b=3∧c=6 ⇒ f(x)=−3x^3 +3x^2 +6x=−3x(x−2)(x+1) But with f(−1)=f(0)=f(2)=0 we directly get f(x)=a(x+1)x(x−2) f(1)=6 ⇒ a×2×1×(−1)=6 ⇒ a=−3 ⇒ f(x)=−3x(x−2)(x+1)

$$\mathrm{Usually}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{polynome} \\ $$$$\mathrm{of}\:\mathrm{minimal}\:\mathrm{degree}.\:\mathrm{With}\:{n}\:\mathrm{equations}\:\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:{n}−\mathrm{1}\:\Rightarrow\:\mathrm{with}\:\mathrm{4} \\ $$$$\mathrm{equations}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{3}. \\ $$$$\mathrm{The}\:\mathrm{usual}\:\mathrm{method}\:\mathrm{is}\:\mathrm{this}: \\ $$$$ \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}={f}\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{d}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$−{a}+{b}−{c}=\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{8}{a}+\mathrm{4}{b}+\mathrm{2}{c}=\mathrm{0}\:\left(\mathrm{3}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{6} \\ $$$${a}+{b}+{c}=\mathrm{6}\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:{b}={a}+{c} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{12}{a}+\mathrm{6}{c}=\mathrm{0}\:\Rightarrow\:{c}=−\mathrm{2}{a}\:\Rightarrow\:{b}=−{a} \\ $$$$\left(\mathrm{4}\right)\:−\mathrm{2}{a}=\mathrm{6}\:\Rightarrow\:{a}=−\mathrm{3}\:\Rightarrow\:{b}=\mathrm{3}\wedge{c}=\mathrm{6} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=−\mathrm{3}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}=−\mathrm{3}{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{with}\:{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)=\mathrm{0}\:\mathrm{we}\:\mathrm{directly} \\ $$$$\mathrm{get} \\ $$$${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{2}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{6}\:\Rightarrow\:{a}×\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)=\mathrm{6}\:\Rightarrow\:{a}=−\mathrm{3} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=−\mathrm{3}{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$

Commented by mathlove last updated on 04/Jan/24

thanks

$${thanks} \\ $$

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