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G''(x) = − e−x − (e−x − xe−x) ∴ g''(x) = − 2e−x xe−x Similarly the third derivative g(3)(x) = 2e−x e−x − xe−x ∴ g(3)(x) = 3e−x − xe−x So it looks like clear pattern is forming, but let us just check by looking at the fourth derivative;Individuals applying for licensure for professions that require !Let F = {0,e,a,b} with addition and multiplication defined by the tables below 0 e a b

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Click on a word in the word list when you've found it This will gray it out and help you remember that you've found itNote that (f B g)(x) ≠ (g B f)(x) This means that, unlike multiplication or addition, composition of functions is not a commutative operation The following example will demonstrate how to evaluate a composition for a given value Example 6 Find (f B g)(3) and (g B f)(3) if f ( x ) = x 2 and g ( x ) = 4 – x2 Solution Step 1 Find (fLearn how to solve f(g(x)) by replacing the x found in the outside function f(x) by g(x)

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Defined as ( mn )˘ mn 2 and g(m,n)˘(m¯1,m¯n)Findtheformulasforg– f and f –g Note g– f ( m,n )˘)) mn 2¯1 Thus g– f ( m,n )˘ mn¯1 2 Note f– g (m,n;f n(x)g is again a continuous function on E Solution For n= 2, use part(a) to write g(x) = (f 1(x) f 2(x)) jf 1(x) f 2(x)j 2 Since f 1 f 2 and jf 1 f 2jare continuous, it follows that gis also continuous For n>2 and k= 2;3;A Z i b k v x m x b k l _ f g _ g b y d Z h j f Z, _ v f _ g g h a j _ r _ g b y e Z ^ _ e v p Z \ l h j k d \ <

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