That is a minimum value, and then we'll be all set. And to do that, weīusiness, figure out where our derivative isĮither undefined or 0, and then just make sure that Need to figure out where this hits a minimum value. Of our combined area based on where we make the cut. Therefore, f(1) fully meets all the requirements of the Second Derivative Test, thus it must be an inflection point. (Note: ε means an arbitrarily small number). You should prove that f(1) is an inflection point with the Second Derivative test:į''(1) = 0, a necessary condition for an inflection point.į''(1-ε) is positive AND f''(1+ε) is negative, a sufficient condition for an inflection point. Thus, the absolute minimum over the given interval must be either f(-3) or f(3), and f(3) is less than f(-3). You can show that f(3) is the absolute min on the given interval because the function is continuous, reaches a max along the interval, has no other local extrema, and f(x) is decreasing as x increases for the interval -1 < x ≤ 3. Thus, f(-1) must be the absolute max over the given interval. In other words, since f(x) is increasing over -3 Thus, it must be the largest value of f(x) over the interval. You can show that x=-1 is an absolute max because it is the only local max, the function is continuous, and there are no other extrema on the given interval. Since you had a defined, closed interval, the largest value is the absolute max FOR THAT INTERVAL.
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