Science:Math Exam Resources/Courses/MATH100/December 2019/Question 6
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Question 6 

Find, with proof, the coordinates of the global maximum and global minimum of the function on the closed interval . 
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 

What are some qualities which a function must have at a point in its domain in order for to be a local maximum or minimum? Note: this will involve the the derivative, but do not forget potential other points. 
Hint 2 

Notice that the domain of our function is a closed interval. So, besides critical points, be sure to also check the endpoints of the domain of our function. 
Hint 3 

Since this is asking for a proof, try to give explanations for your calculations. This helps demonstrate to the reader the reasoning behind your proof. 
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. We are asked to find the global maximum and global minimum of the function on the closed interval . Because the domain of our function is a closed interval, we need to check the lefthand and righthand endpoints, as well as critical points. These are the points in the domain of where , or where the derivative of fails to exist. Let us first identify critical points. First, since , the function is welldefined (i.e. makes sense) and differentiable. Moreover, utilizing the product and power rules, we see that: . We now assume that is a real number, with , and . Then, . So, we have is the only critical point of the function in the interval . Now, let us test our three points by plugging them into our function . We have:
From the above, on the interval , has global minimum , and global maximum at the point . This is because . 