Function Problem 6

Calculus Level 2

Let f ( x ) = ( x + 1 ) 2 − 1 , x ≥ − 1 f(x)=(x+1)^2-1,x\ge{-1} ,then the set { x : f ( x ) = f − 1 ( x ) } \{x:f(x)=f^{-1}(x)\} is equal to:-

{ − 1 , 1 } \{-1,1\} { 0 , 1 } \{0,1\} None Of These \text{None Of These} { 0 , − 1 } \{0,-1\}

Parag Zode by Parag Zode , 24, India

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1 solution

Parag Zode
Dec 29, 2014

We just see that according to condition , f ( x ) = f − 1 ( x ) = x f(x)=f^{-1}(x)=x ⟹ ( x + 1 ) 2 − 1 = x \implies(x+1)^2-1=x ⟹ x + 1 = 0 \implies x+1=0 or x = 0 x=0 and thus we conclude that the set is equal to x = { 0 , − 1 } x=\{0,-1\}

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