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Can someone explain problems 2B7 a & b. I've reviewed the answer and it doesn't make sense to me. I'm missing a concept here.
http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/unit2applicationsofdifferentiation/partaapproximationandcurvesketching/problemset3/MIT18_01SC_pset2prb.pdf
 one year ago
 one year ago
Can someone explain problems 2B7 a & b. I've reviewed the answer and it doesn't make sense to me. I'm missing a concept here. http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/unit2applicationsofdifferentiation/partaapproximationandcurvesketching/problemset3/MIT18_01SC_pset2prb.pdf
 one year ago
 one year ago

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WaynexBest ResponseYou've already chosen the best response.1
If f(x) is increasing at a given point, then the slope at that point has two forms it could be in. Either y and x are both positive, or y and x are both negative. If y and x are both negative, then the negatives on the top and bottom of the fraction cancel, and the slope is still positive. Therefore, when f(x) is increasing, the slope is always positive (or zero, as you will see shortly, hopefully). The derivative gives us a method to calculate the slope at a given point, hence, if f(x) is increasing, the derivative at that given point is positive. Since the limit of a continuous function is the same as the derivative at a given point, you can use the limit to prove this. For part b, x to the third power has an inflection point, a point were the function is increasing, and then for one instance goes horizontal and has slope of zero. It's just one point. Then it increases again. The function is said to be increasing at this inflection point because it does not decrease. It simply goes from increasing, to horizontal for one instantaneous point, and then increasing again.
 one year ago

newhite1Best ResponseYou've already chosen the best response.0
Thank you for your explanation. When I revisited the problem with your explanation, it help connected the dots.
 one year ago
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