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Math Bridging Course Tutorial 3 Chris TC Wong 30/8/2012 1/9/2012
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Review on Maximum and Minimum Concept
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Existence of Maximum and Minimum For this function, does global Maximum exists on… [-5,5] (-5,5)
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Extreme value theorem If a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain its maximum and minimum value, each at least once. That is, there exist numbers c and d in [a,b] such that: f ( c ) <= f ( x ) < f ( d ) for any x in [a,b] (http://en.wikipedia.org/wiki/Extreme_value_theorem)http://en.wikipedia.org/wiki/Extreme_value_theorem
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Strange things Closed? Bounded? Continuous function? V.s. Open (Supremum and Infimum) Unbounded (what is infinity ?) Not continuous function (Where is the “break point”?)
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Assume things are nice The function is differentiable. (Hence also continuous) i.e. first derivative exists. First derivative test. Nicer : the function is twice differentiable i.e. second derivative exists. Second derivative test. Very Nice : the function is “smooth” i.e. Derivative of any order exists
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First derivative test X<-1X=-1-1<X<1X=1X>1 f(x)↗2/3↘-2/3↗ f’(x)+0-0+
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Caution :
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Algorithm Read carefully about the function Differentiate the function Finding local max/min Compute function value on Boundary points Compute function value on non-differentiable points Return max{f(BoundaryPts),f(non-d-able-pts),localMaxs} and min{f(BoundaryPts),f(non-d-able-pts),localMins}
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Second Derivative test X<-1x=-1-1<x<0X=00<X<1X=1X>1 f(x)↗2/3↘0↘-2/3↗ f’(x)+0---0+ f‘’(x)---0+++
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Who cares about point of inflexion?
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Exercises
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Q&A
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