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Question Number 204671 by mr W last updated on 25/Feb/24
if f(f(x))=x^2 −3x+4, find f(1)=?
iff(f(x))=x2−3x+4,findf(1)=?
Answered by witcher3 last updated on 25/Feb/24
f(f(x))=x^2 −3x+4 f(f(1))=2 f((f(2))=2 ⇒f(f(f(2))=f(2)=f^2 (2)−3f(2)+4⇒f(2)=2 f(f(f(1))=f(2)=2 ⇒f(1)^2 −3f(1)+4=2⇒f^2 (1)−3f(1)+2=0 ⇒f(1)∈{1,2} f(f(1))=2⇒f(1)=2 impossible ⇒f(1)=2
f(f(x))=x2−3x+4f(f(1))=2f((f(2))=2⇒f(f(f(2))=f(2)=f2(2)−3f(2)+4⇒f(2)=2f(f(f(1))=f(2)=2⇒f(1)2−3f(1)+4=2⇒f2(1)−3f(1)+2=0⇒f(1)∈{1,2}f(f(1))=2⇒f(1)=2impossible⇒f(1)=2
Commented by mr W last updated on 25/Feb/24
great!
great!
Answered by A5T last updated on 25/Feb/24
f(x)=f(y)⇒f(f(x))=f(f(y))⇒x^2 −3x=y^2 −3y ⇒(x−y)(x+y)=3(x−y)⇒x=y or x=3−y ⇒f(x)=f(3−x)⇒f(1)=f(2) f(f(2))=2 f(f(f(2)))=f(2)=[f(2)]^2 −3f(2)+4⇒[f(2)−2]^2 =0 ⇒f(2)=2=f(1)
f(x)=f(y)⇒f(f(x))=f(f(y))⇒x2−3x=y2−3y⇒(x−y)(x+y)=3(x−y)⇒x=yorx=3−y⇒f(x)=f(3−x)⇒f(1)=f(2)f(f(2))=2f(f(f(2)))=f(2)=[f(2)]2−3f(2)+4⇒[f(2)−2]2=0⇒f(2)=2=f(1)
Commented by mr W last updated on 25/Feb/24
great!
great!
Answered by es last updated on 25/Feb/24
f(x)=f^(−1) (x^2 −3x+4)→ f(1)=f^(−1) (2) f^(−1) (x)=((3±(√(9−4(4−x))))/2) f^(−1) (2)=((3±1)/2)=f(1)= { (2),(1) :}
f(x)=f−1(x2−3x+4)→f(1)=f−1(2)f−1(x)=3±9−4(4−x)2f−1(2)=3±12=f(1)={21

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