Asked by Rob
prove that 5x - 7 - sin3x = 0 has at least one zero and prove that it has exactly one real zero.
How am I supposed to show my work for this? I don't really understand how to show my work for the IVT. Without that I can't do any work for the MVT.
How am I supposed to show my work for this? I don't really understand how to show my work for the IVT. Without that I can't do any work for the MVT.
Answers
Answered by
drwls
Draw graphs of 5x - 7 and sin3x and see how many times they intersect. They only cross once, at about x = 1.27. That can be clearly seen from the two graphs.
I believe in doing things the easy way. I don't see how either the mean or intermediate value theorems will be useful here, but math is not my specialty.
5x - 7 only has values that are in the +/- 1 range of sin 3x when x is between 1.2 and 1.6. Perhaps that will help with a more formal proof.
You only have to consider the function
5x - 7 - sin3x between x = 1.2 and x = 1.6.
In that interval, it goes from -0.5575 to 1.996.
The slope remains positive so it can only equal zero once.
I believe in doing things the easy way. I don't see how either the mean or intermediate value theorems will be useful here, but math is not my specialty.
5x - 7 only has values that are in the +/- 1 range of sin 3x when x is between 1.2 and 1.6. Perhaps that will help with a more formal proof.
You only have to consider the function
5x - 7 - sin3x between x = 1.2 and x = 1.6.
In that interval, it goes from -0.5575 to 1.996.
The slope remains positive so it can only equal zero once.
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