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Functions: Give an appropriate domain for #g(x)# which enables #g^-1(x)# to exist?

#g(x) = 1/(x+7)^2#

#g^-1(x) = -1/sqrt(x) - 7#

The questions asks to find a domain for #g(x)# such that it also enables #g^-1(x)# to exist.

My first answer was:

#{y:y>=0}#

but apparently this is wrong. Any help would be much appreciated!

Answer 1

The domain of #g(x)# is #x in RR | x!=-7#

The domain for #g(x)# is #x in RR | x !=-7# because #x = -7# will cause division by 0 and this is not allowed.

Furthermore, the range for #g(x)# is #y >0#

The range of the function is the domain of the inverse function, and the domain of the function is the range of the inverse function.

Please observe that there is no value of x that will make #g^-1(x) = -7# and #x >0# is the domain for #g^-1(x)#

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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.

Functions: Give an appropriate domain for #g(x)# which enables #g^-1(x)# to exist?

#g(x) = 1/(x+7)^2# #g^-1(x) = -1/sqrt(x) - 7# The questions asks to find a domain for #g(x)# such that it also enables #g^-1(x)# to exist. My first answer was: #{y:y>=0}# but apparently this is wrong. Any help would be much appreciated!

#g(x) = 1/(x+7)^2#

#g^-1(x) = -1/sqrt(x) - 7#

The questions asks to find a domain for #g(x)# such that it also enables #g^-1(x)# to exist.

My first answer was:

#{y:y>=0}#

but apparently this is wrong. Any help would be much appreciated!