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Question
1.
Let f be integrable on (-infty,+infty). Suppose a function F is defined by F(x)=integral from -infty to x of f(t)dt. Prove that F is continuous on (-infty,+infty).
2. Using continuity, find an integer n so that the equation 2x cos(x)=(x+2)^2 has a solution on [n,n+1]. (There are 2 possible answers.) Prove that your choice for n is correct.
1. Since the derivative is the opposite of the integral,
anything that can be integrated can therefore be differentiated.
Since it can be differentiated, we know it is contiuous.
2. Construct a triangle that has a near side of (x+2)²
and a hypoteneuse of 2x. Using this, the angle can be found.
I don't know whether the equation is in degrees or radians.
Either way, an x too large and an x to small can be found between 0 and pi/4 radians or 0 and 90 degrees. Find the angle that is halfway inbetweeen and match it up with the one that is on the opposite side of the solution. This is called the bisection method
and can be carried to be as accurate as needed.
For example, suppose you were trying to solve 0 = x² - 200.
You could use, say x1 = 10 and x2 = 20.
That's 10²=100 is too small and 20²=400 is too large.
Try x = (x1+x2)/2 - 15. Now 15²=225, which is still too large.
Since 20 is also too large, replace the 20 with the 15.
Now take the lower limit 10 and the upper limit 15.
The average is (10+15)/2 = 12.5.
When you look at 12.5², you get 156.25. We are looking for the answer to be 200, so that's too low, therefore it replaces the 10. Are two numbers are now 12.5 and 15.
The average of 12.5 and 15 is 13.75. 13.75²=189.0625, which is also low. This then replaces the 12.5 and our new numbers are
13.75 and 15. The average of these two is 14.375. Now 14.375² is
206.46, which is too high. Therefore the new numbers are 13.75 and 14.375.
Got the idea? Do the same thing with arccos(x+2)²/2x.
Note that making the far side negative will position the angle in a different quadrant. This should give the other solution.
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Scott A Wilson
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