<p>I am unsure of how to proceed with these problems. Below each question, I provided the work that I have so far (I don’t know if it’s right or how to continue).</p>
<li>If f(x) = x^2 + 10 sin[x], show that there is a number c such that f(c) = 1000.</li>
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<p>okay, well you're definately on the right track for 45
I would go and change it to a^2+10sin[a]-1000<0
some teachers would allow you to say that a^2+10sin[a]-1000<0<b^2+10sin[a]-1000 therefore there is a zero between the [a,b].
If you have a really strict teacher, she'd make you try and figure out a and b, but that's really difficult.</p>
<p>Also, I would try and prove it w/ a number as a and another (greater) as b to prove it algebraically. That's as much help I could give you. I would try and do the same with your other question.</p>
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<li>Try to find a number that yields a negative number and another one that yields a positive number. By the Intermediate Value Theorem, since 0 is between positive and negative numbers, there has to be a root.</li>
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<ol>
<li>If f(x) = x^2 + 10 sin[x], show that there is a number c such that f(c) = 1000.</li>
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<p>I'll start by making the obvious note that f(x) is continuous. What you're trying to do is find x1, x2 such that f(x1) < 1000 and f(x2) > 1000. This shouldn't be too hard to do since |max(10 sin[x])| = 10. However, in the following method, I'll be using a multiple of pi as x1 and x2 so that sin(x1) = sin(x2) = 0. So you just need to worry about x^2 being less than and, later, greater than 1000.</p>
<p>Solution: Choose x1 = 0. Then f(x1) = 0 < 1000. To find x2 > 1000, find k such that (pi<em>k)^2 > 1000, since sin(pi</em>k) = 0. Finding such a k is easy, as you can just pick a large number. For example, choose k = 100. Then choose x2 = pi<em>k = 100pi. Then f(x2) = 10000</em>pi^2 > 1000. The result follows from the intermediate value theorem.</p>
<p>Note: In your proof, you must mention that f(x) is continuous. If it is not, then making a conclusion from the intermediate value theorem is impossible.</p>
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<li><p>f(x) is a continuous function everywhere, so we can apply the IMT with any appropriate interval. For example, a=0, b=100. f(0) = 0, f(100) = 10,000 + 10sin(100) >= 9990. Since 1000 is in the interval [ f(a), f(b) ] there must be a c in [a,b] such that f(c) = 1000.</p></li>
<li><p>Same argument; you can use [a,b] = [0,2]. f(0) = -1, f(2)>= 7.</p></li>
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