Science:Math Exam Resources/Courses/MATH102/December 2014/Question A 01
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Question A 01 

Suppose that is the inverse function of The tangent line to at is where and are constants. Which of the following is the equation of a tangent line to 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Recall that the tangent line to at is given by (Why? Consider the slope of the tangent line and the fact that it must pass through ...) 
Hint 2 

Following the first hint, we see that and Given this information, what point must lie on the graph of (the inverse of )? 
Hint 3 

implies that since Hence the point must lie on the graph of What is 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 1 

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Please rate my easiness! It's quick and helps everyone guide their studies. As the tangent line to at is given by it is clear that and Since Now recall* that so Since the tangent line to at is given by the line is tangent to the graph of as desired. * If you do not recall this, observe that by the chain rule, But and so 
Solution 2 

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Please rate my easiness! It's quick and helps everyone guide their studies. Since the graph of , the inverse of , is simply the graph of reflected across the line a tangent line to the graph of can be found by simply reflecting a tangent line to the graph of across the same line. In other words, 'a tangent to the inverse is the inverse of a tangent'. Algebraically, we simply interchange and in the equation thereby obtaining Solving for we arrive at 