Answer:
Given f(x)=x
Given f(x)=x 2
Given f(x)=x 2 −2x+4=(x−1)
Given f(x)=x 2 −2x+4=(x−1) 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1)
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1)
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2)
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2 ⇒x−2=±x
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2 ⇒x−2=±x⇒x−2=x or x−2=−x
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2 ⇒x−2=±x⇒x−2=x or x−2=−x⇒−2=0 or x+x=2⇒x=1
Given f(x)=x 2 −2x+4=(x−1) 2 +3It is also given that, f(x−1)=f(x+1)⇒(x−1−1) 2 +3=(x+1−1) 2 +3⇒(x−2) 2 =x 2 ⇒x−2=±x⇒x−2=x or x−2=−x⇒−2=0 or x+x=2⇒x=1So the only solution is {1}
