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Let f:R to [1,oo),f(x)=x^(2)-4x+5. Then ...

Let `f:R to [1,oo),f(x)=x^(2)-4x+5`. Then find the largest possible intervals for which `f^(-1)(x)` is defined and find corresponding `f^(-1)(x).`

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We have `f:R to [1,oo),f(x)=x^(2)-4x+5`
or `f(x) =(x-2)^ (2)+1.`
Graph of the function is as shown in the following figure.

Clearly f(x) is many-one function.
However, in the interval `(-oo,2] or [2,oo), f(x)` is one-one.
For ` f:(-oo,2] to [1,oo),y=(x-2)^(2) +1`
` :. (x-2)^(2)=y-1`
` implies x-2=sqrt(y-1)`
`implies x=2-sqrt(y-1)`
Thus `f^(-1):[1,oo) to (-oo,2],f^(-1)(x)=2-sqrt(x-1)`
For `f:[2,oo) to [1,oo),y=(x-2)^(2)+1`
`implies x=2 +sqrt(y-1)`
Thus ` f^(-1):[1,oo) to [2,oo),f^(-1)(x)=sqrt(x-1)+2`
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