Function and their inverse

Algebra Level 5

f ( x ) = { − x + 1 , x ≤ 0 − ( x − 1 ) 2 , x ≥ 1 \large f(x)=\begin{cases} -x+1, \quad \ \ \ \ x \leq 0 \\ -(x-1)^2, \quad x \geq 1 \end{cases}

Define the piecewise function of f ( x ) f(x) above, find the number of solution(s) of the equation f ( x ) − f − 1 ( x ) = 0 f(x)-f^{-1}(x)=0 .

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2 0 1 3 infinitely many 4

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Rohit Ner
May 15, 2015

f ( x ) − f − 1 ( x ) = 0 f(x)-f^{-1}(x)=0

⇒ f ( f ( x ) ) = x \Rightarrow f\left( f\left( x \right) \right) =x

Case 1: x ≤ 0 x\le 0

f ( x ) = − x + 1 f\left( x \right) =-x+1 . . . ≥ 1 \ge 1

f ( f ( x ) ) = − ( f ( x ) − 1 ) 2 f\left( f\left( x \right) \right) =-{ \left( f\left( x \right) -1 \right) }^{ 2 }

= − ( − x + 1 + 1 ) 2 = − ( − x ) 2 = − x 2 =-{ \left( -x+1+1 \right) }^{ 2 }=-{ \left( -x \right) }^{ 2 }=-{ x }^{ 2 }

⇒ x = − x 2 ⇒ x = 0 , − 1 \Rightarrow x=-{ x }^{ 2 }\Rightarrow x=0,-1

Case 2: x ≥ 1 x\ge 1

f ( x ) = − ( x − 1 ) 2 f\left( x \right) =-{ \left( x-1 \right) }^{ 2 } . . . ≤ 0 \le 0

f ( f ( x ) ) = − f ( x ) + 1 f\left( f\left( x \right) \right) =-f\left( x \right) +1

= ( x − 1 ) 2 + 1 ={ \left( x-1 \right) }^{ 2 }+1

⇒ x = ( x − 1 ) 2 + 1 ⇒ x = 1 , 2 \Rightarrow x={ \left( x-1 \right) }^{ 2 }+1\Rightarrow x=1,2

∴ x = − 1 , 0 , 1 , 2 \therefore x=-1,0,1,2

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice. There is a Typo.

= − ( − x + 1 − 1 ) 2 = − ( − x ) 2 = − x 2 =-{ \left( -x+1{\Huge \color{#D61F06}{-}}1 \right) }^{ 2 }=-{ \left( -x \right) }^{ 2 }=-{ x }^{ 2 }

Niranjan Khanderia - 5 years, 11 months ago

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This one was an eye opener to me @Sandeep Bhardwaj . Till now I used to think that the only solution would lie on y=x or else infinite solutions, when the function is the inverse of itself

Thanks

Mayank Singh - 5 years, 7 months ago

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My pleasure! The same was my intention to post this problem. ⌣ ¨ \ddot \smile

Sandeep Bhardwaj - 5 years, 7 months ago

In region x ⩽ 0 x\leqslant0 f − 1 ( x ) = f ( x ) \quad\quad f^{-1}(x)=f(x) . f ( − x + 1 ) = − ( − x + 1 ) + 1 = x − 1 + 1 = x f(-x+1)=-(-x+1)+1=x-1+1=x , so why are there not infinitely many solutions.

Miloje Đukanović - 5 years, 8 months ago

why cant we do this by graph

Shubham Rustagi - 4 years, 6 months ago

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