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How do you find the inverse of #f(x) = x^2 +x#?

Answer 1

#f^-1(x)=-1/2+-sqrt(x+1/4)#

To find an inverse, switch x and y (or f(x)).

For simplicity I will rewrite f(x) as y... #y=x^2+x# switch... #x=y^2+y# now solve for y It looks like we need to complete the square... #x+1/4=y^2+y+1/4# #x+1/4=(y+1/2)^2# #+-sqrt(x+1/4)=y+1/2# Don't forget the #+-#! #-1/2+-sqrt(x+1/4)=y# Don't forget the #+-#!

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Answer 2

Sadie Alexander

To find the inverse of the function ( f(x) = x^2 + x ), follow these steps:

Replace ( f(x) ) with ( y ).
Swap the roles of ( x ) and ( y ): ( x = y^2 + y ).
Solve the equation obtained in step 2 for ( y ).
Replace ( y ) with ( f^{-1}(x) ) to express the inverse function.

Here's the process:

[ x = y^2 + y ]

[ x = y(y + 1) ]

[ y(y + 1) = x ]

[ y^2 + y - x = 0 ]

Using the quadratic formula:

[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

where ( a = 1 ), ( b = 1 ), and ( c = -x ), we have:

[ y = \frac{-1 \pm \sqrt{1 + 4x}}{2} ]

So, the inverse function ( f^{-1}(x) ) is:

[ f^{-1}(x) = \frac{-1 \pm \sqrt{1 + 4x}}{2} ]

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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.