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How do you find the domain and range of #h(x)=ln(x-6)#?

Answer 1

The answers are: #D(6,+oo)# and #R(-oo,+oo)#.

The domain of the function #y=lnf(x)# is: #f(x)>0#.

So:

#x-6>0rArrx>6# or we can write: #D=(6,+oo)#

The range of a function is the domain of the inverse function. The inverse function of the logarithmic function is the exponential function.

So (using the method to find the inverse function, that is: exchange #x# with #y# and finding #y#):

#y=ln(x-6)rArrx=ln(y-6)rArre^x=y-6rArry=e^x+6#,

that has domain #(-oo,+oo)#.

The function is:

graph{ln(x-6) [-2, 15, -5, 5]}

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Answer 2

Adam Adkins

To find the domain of ( h(x) = \ln(x - 6) ), we need to consider the values that ( x ) can take such that the natural logarithm function is defined. Since the natural logarithm is only defined for positive real numbers, we set the argument ( x - 6 ) greater than zero and solve for ( x ):

[ x - 6 > 0 ]

[ x > 6 ]

So, the domain of ( h(x) ) is all real numbers greater than 6, expressed as ( x \in (6, \infty) ).

To find the range of ( h(x) ), we need to consider the possible values that the natural logarithm function can output. The range of the natural logarithm function ( \ln(x) ) is all real numbers, which means that ( h(x) ) can take any real value.

Therefore, the range of ( h(x) = \ln(x - 6) ) is all real numbers, expressed as ( h(x) \in (-\infty, \infty) ).

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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.