1. Math Bridging Course Tutorial 3 Chris TC Wong 30/8/2012 1/9/2012

2. Review on Maximum and Minimum Concept

3. Existence of Maximum and Minimum For this function, does global Maximum exists on… [-5,5] (-5,5)

4. 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

5. Strange things Closed? Bounded? Continuous function? V.s. Open (Supremum and Infimum) Unbounded (what is infinity ?) Not continuous function (Where is the “break point”?)

6. 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

7. First derivative test X<-1X=-1-1<X<1X=1X>1 f(x)↗2/3↘-2/3↗ f’(x)+0-0+

8. Caution :

9. 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}

10. 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+++

11. Who cares about point of inflexion?

12. Exercises

13. Q&A