How do you find the inverse of #g(x) = x^2 + 4x + 3 # and is it a function?

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Answer 1 · 2018-04-05T15:48:44

See below.

To find the inverse function, we need to express #x# as a function of #y#:

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

Substitute:

#y=x#

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

Subtract #x#:

#y^2+4y+3-x=0#

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

#y=(-4+-sqrt(16-12+4x))/2#

#y=(-4+-sqrt(4+4x))/2#

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

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

#:.#

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

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

If we look at the inverses, remembering that #x# is the range of the function, we can see that, for:

#sqrt(1+x)#

#1+x>=0#

#x>=-1#

This means that:

#x^2+4x+3>=-1#

Solving:

#x^2+4x+4>=0#

Factor:

#(x+2)^2>=0#

So the inverses are defined for the domain of the function.

They are both functions if we take #sqrt(1+x)# as meaning the principal root.