Posted by **Rob** on Friday, December 16, 2011 at 1:46am.

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.

- calculus -
**drwls**, Friday, December 16, 2011 at 2:23am
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.

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