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How do you find #f^-1(x)# given #f(x)=x^2-4x+3#?

Answer 1

Serenity Jones

#f^-1(x) = 2 +- sqrt(x + 1)#

We switch the x-y values by the definition of the inverse function (a reflection over the line #y =x#).

#x = y^2 - 4y + 3#

We now complete the square in order to solve for #y#.

#x = 1(y^2 - 4y + 4 - 4) + 3#

#x = 1(y - 2)^2 - 4 + 3#

#x= 1(y - 2)^2 - 1#

#x= (y - 2)^2 - 1#

Solve for #y#:

#x + 1 = (y - 2)^2#

#+-sqrt(x +1) = y - 2#

#2 +-sqrt(x + 1) = y#

Hopefully this helps!

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Answer 2

Greyson Brown

To find the inverse function ( f^{-1}(x) ) given ( f(x) = x^2 - 4x + 3 ), follow these steps:

Replace ( f(x) ) with ( y ): ( y = x^2 - 4x + 3 ).
Swap ( x ) and ( y ): ( x = y^2 - 4y + 3 ).
Rearrange the equation to solve for ( y ).

[ x = y^2 - 4y + 3 ] [ y^2 - 4y = x - 3 ] [ y^2 - 4y + 4 = x - 3 + 4 ] [ (y - 2)^2 = x + 1 ]

Take the square root of both sides.

[ y - 2 = \pm \sqrt{x + 1} ]

Solve for ( y ).

[ y = 2 \pm \sqrt{x + 1} ]

So, ( f^{-1}(x) = 2 + \sqrt{x + 1} ) or ( f^{-1}(x) = 2 - \sqrt{x + 1} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.