Question
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Use the given function to answer parts (a) through (f) below.

© From the graph, determine the range and any asymptotes of .
The range of is .
(Type your answer in interval notation.)
Determine the vertical asymptote of . Select the correct choice below and, if necessary, fill in the answer box to
complete your choice.
B. The vertical asymptote of is . There is no vertical asymptote.
(Simplify your answer.)
(d) Find , the inverse of .

(Simplify your answer.)
(e) Find the domain and range of .
The domain of is .
(Type your answer in interval notation.)

Ask by Alexander Savage. in the United States
Mar 26,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

© The range of is . The vertical asymptote of is .
(d) The inverse function is .
(e) The domain of is .

Solution

© The function is given by
Since the logarithm is defined for positive arguments, we have
Thus, the domain of is . For logarithmic functions of the form , the horizontal output covers all real numbers. Therefore, the range of is
The vertical asymptote occurs where the argument of the logarithm is zero. Setting
we see that the vertical asymptote is at

(d) To find the inverse function , start with
Interchange and to obtain
Subtract 2 from both sides:
Now rewrite the logarithmic equation in exponential form (assuming base 10):
Finally, solve for :
Thus, the inverse function is

(e) The domain of is the range of , and the range of is the domain of . Since
  • The range of is , the domain of is
  • The domain of is , so the range of is
Thus, the domain of is and its range is .

Answered by UpStudy AI and reviewed by a Professional Tutor

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Simplify this solution

Mind Expander

The range of the function is indeed because the logarithmic function itself spans all real numbers. The vertical asymptote at represents the point where the function’s value approaches negative infinity as approaches 4 from the right; thus, it’s correct to say that the vertical asymptote is .
For the inverse, when you find , you’re switching and in the original function and solving for . Confirming is great because it reflects the logarithmic nature of the original function. Now, for the domain and range of , since the range of is all real numbers, the domain of is also . So, simply fill in the box with . Enjoy your exploration of these functions!

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