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Finding Absolute Extrema :Now it's time to see our first major application of derivatives. Specified a continuous function, f(x), on an interval [a,b] we desire to find out the absolute extrema of the function. To do this we will requierd many of the ideas which we looked at in the previous section.
Firstly, as we have an interval and we are considering that the function is continuous the Extreme Value Theorem described that we can actually do this. it is a good thing of course. We don't desire to be trying to determine something that may not exist.
Next, we illustrated in the earlier section that absolute extrema can take place at endpoints or at relative extrema. Also, from Fermat's Theorem we know that the list of critical points is also a list of all probable relative extrema. Thus the endpoints along with the list of all critical points will actually be a list of all probable absolute extrema.
Now we just required to recall that the absolute extrema are nothing more than the largest & smallest values which a function will take thus all that we actually required to do is get a list of possible absolute extrema, plug these points into our function and then recognize the largest & smallest values.
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If the points (5, 4) and (x, y) are equidistant from the point (4, 5), prove that x 2 + y 2 - 8x - 10y +39 = 0. Ans : AP = PB AP 2 = PB 2 (5 - 4) 2 + (4 - 5) 2 = (x
express 4:24 as fraction in lowest term
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