Science:Math Exam Resources/Courses/MATH104/December 2014/Question 06
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Question 06 

Let be a function defined on the interval . Show that the graph of on the interval has an inflection point. 
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 

First simplify the function by using the change the base formula for logarithms. 
Hint 2 

Think about what the necessary and sufficient conditions are for there to be an inflection point. 
Hint 3 

Additionally, Intermediate Value Theorem will be very very useful here. 
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 

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Please rate my easiness! It's quick and helps everyone guide their studies. The necessary condition for to have an inflection point is that there exists a in so that . A sufficient condition for to have an inflection point is that changes signs at .
First, let us simplify the function . Using the change of base of logarithms formula given by
Take , , and , we obtain that Thus
Now we take the first and second derivatives of . Recall that if , then . Taking the first derivative of , we get:
Now, we take the second derivative of and find:
Remark next that is a continuous function of on the interval , since both and are continuous functions of on . Now we want to show that there exists a so that . Evaluate gives:
Now evaluate to get: . Since is continuous on , and while , by the Intermediate Value Theorem, there exists a point so that . We have already shown that there is a point to the left and to the right of c with different signs. So indeed the given function has an inflection point on . 