Question-215052 Tinku Tara December 27, 2024 Algebra 0 Comments FacebookTweetPin 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)+1f(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]+1f(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}+1sayk(x)=ax+bf(x)=(x3−6x2+11x−6)(ax+b)−2x2+9x−6f′(x)=a(x3−6x2+11x−6)+(3x2−12x+11)(ax+b)−4x+9f′(1)=1⇒a+b=−2f′(3)=1⇒3a+b=2⇒a=2,b=−4f(x)=(x−1){2(x−2)[(x−3)(x−2)−1]+3}+1orf(x)=2x4−16x3+44x2−47x+18f(4)=22✓ Commented by mr W last updated on 27/Dec/24 Answered by mr W last updated on 27/Dec/24 alternative:sayg(x)=f(x)−x⩾0f(1)−1=g(1)=0f(3)−3=g(3)=0f(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)−2g(x)=(x−1)(x−3)[(x−2)k(x)−2]sayk(x)=ax+bg′(1)=0,g′(3)=0…… Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Solve-for-a-b-c-R-1-a-1-b-c-1-2-1-b-1-c-a-1-3-1-c-1-a-b-1-4-Next Next post: n-0-1-n-2-n-1-2n-1-p-q-p-q-p-q- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.