Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)
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Question 03 (c)
Consider the following function, where is a constant.
For what value(s) of does the equation have at least one solution? Explain why. Also, make sure you state what theorem(s) you may be using to justify your claim and why they apply.
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.
To show that this equation has zero is equivalent to show that the function has intersection with . Observe that the intersection can only happen in the interval .
Apply intermediate value theorem on the interval .
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.
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Before we apply intermediate value theorem, let’s discuss why the intersection can only happen in rather than .
We know from the question that when . Since the exponential function is an increasing function, we have when . On the other hand, is decreasing, and hence so is . This follows that when . This implies that for , we have
(Note that .) In other words, for any point in .
In a similar manner, we can show that when , we have that
The above analysis would be more obvious if you draw the graphs of and .
On the interval , let
(Here, we use on .)
It is time to apply intermediate value theorem, note that is continuous in the interval , and
To make have at least a zero in , by the theorem, we need to make sure that and have opposite signs. i.e., .
Solving this inequality gives us that
Thus, we have at least one solution to (i.e., ) if