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Question-215052




Question Number 215052 by Abdullahrussell last updated on 27/Dec/24
Answered by mr W last updated on 27/Dec/24
f(1)=1 ⇒f(x)=(x−1)g(x)+1  f(2)=(2−1)g(2)+1=4 ⇒g(2)=3 ⇒g(x)=(x−2)h(x)+3  ⇒f(x)=(x−1)[(x−2)h(x)+3]+1  f(3)=(3−1)[(3−2)h(3)+3]+1=3 ⇒h(3)=−2 ⇒h(x)=(x−3)k(x)−2  ⇒f(x)=(x−1){(x−2)[(x−3)k(x)−2]+3}+1  say k(x)=ax+b  f(x)=(x^3 −6x^2 +11x−6)(ax+b)−2x^2 +9x−6  f′(x)=a(x^3 −6x^2 +11x−6)+(3x^2 −12x+11)(ax+b)−4x+9  f′(1)=1 ⇒a+b=−2  f′(3)=1 ⇒3a+b=2  ⇒a=2, b=−4  f(x)=(x−1){2(x−2)[(x−3)(x−2)−1]+3}+1  or  f(x)=2x^4 −16x^3 +44x^2 −47x+18  f(4)=22 ✓
f(1)=1f(x)=(x1)g(x)+1f(2)=(21)g(2)+1=4g(2)=3g(x)=(x2)h(x)+3f(x)=(x1)[(x2)h(x)+3]+1f(3)=(31)[(32)h(3)+3]+1=3h(3)=2h(x)=(x3)k(x)2f(x)=(x1){(x2)[(x3)k(x)2]+3}+1sayk(x)=ax+bf(x)=(x36x2+11x6)(ax+b)2x2+9x6f(x)=a(x36x2+11x6)+(3x212x+11)(ax+b)4x+9f(1)=1a+b=2f(3)=13a+b=2a=2,b=4f(x)=(x1){2(x2)[(x3)(x2)1]+3}+1orf(x)=2x416x3+44x247x+18f(4)=22
Commented by mr W last updated on 27/Dec/24
Answered by mr W last updated on 27/Dec/24
alternative:  say g(x)=f(x)−x≥0  f(1)−1=g(1)=0  f(3)−3=g(3)=0  f(2)−2=g(2)=4−2=2  ⇒g(x)=(x−1)(x−3)h(x)  g(2)=(2−1)(2−3)h(2)=−h(2)=2  ⇒h(x)=(x−2)k(x)−2  g(x)=(x−1)(x−3)[(x−2)k(x)−2]  say k(x)=ax+b  g′(1)=0, g′(3)=0   ......
alternative:sayg(x)=f(x)x0f(1)1=g(1)=0f(3)3=g(3)=0f(2)2=g(2)=42=2g(x)=(x1)(x3)h(x)g(2)=(21)(23)h(2)=h(2)=2h(x)=(x2)k(x)2g(x)=(x1)(x3)[(x2)k(x)2]sayk(x)=ax+bg(1)=0,g(3)=0

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