009A Sample Final 2, Problem 7
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Show that the equation has exactly one real root.
Foundations: |
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1. Intermediate Value Theorem |
If is continuous on a closed interval and is any number |
between and then there is at least one number in the closed interval such that |
2. Mean Value Theorem |
Suppose is a function that satisfies the following: |
is continuous on the closed interval |
is differentiable on the open interval |
Then, there is a number such that and |
Solution:
Step 1: |
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First, we note that |
Also, |
Since and |
there exists with such that |
by the Intermediate Value Theorem. |
Hence, has at least one zero. |
Step 2: |
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Suppose that has more than one zero. |
So, there exist such that |
Then, by the Mean Value Theorem, there exists with such that |
We have |
Since |
Therefore, it is impossible for Hence, has at most one zero. |
Final Answer: |
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See solution above. |