Elementary Functions (f + g)(x) = x2 + (x + 1);

Transcription

1 The alebra o unctions Given two unctions, say (x) = and (x) = x +, we can, in obvious ways, add, subtract, multiply and divide these unctions For example, the unction + is deined simply by ( + )(x) = + (x + ); Part, Functions Lecture 5a, the unction is deined simply by ( )(x) = (x + ) Dr Ken W Smith Similarly ( )(x) = ( ) (x + ) and / 0 The alebra o unctions Given two unctions, say (x) = and (x) = x +, we can divide by A more important operation between unctions is the operation o / 0 unction composition I is a unction rom X into Y and is a unction rom Y into Z then is a unction rom X into Z deined by irst allowin to map elements o X into Y and then allowin elements o Y to be mapped by into Z or divide by : I we are dividin one unction by another, the quotient is not deined whenever the denominator is zero So does not have zero in its domain Similarly we cannot plu in x = to the quotient 3 / 0 4 / 0

2 Notice that in, is the irst unction involved while is the second! We read unction notation ( )(x) rom riht to let In the example below, maps the elements o X as ollows: View a unction as a machine, takin inputs and eneratin outputs The composition o two unctions will be a sequence o unction machines: c # d!! 5 / 0 6 / 0 I the codomain o the unction is the same as the domain o the unction, then we can compose irst then to create ( ) Or we can compose irst then to create ( ) But the order o composition is important! The unction ( ) is probably not the same unction as ( )! For example, i (x) = x and (x) = x + then (as done above) we have ( )(x) = x + Suppose (x) = x and (x) = x + The unction ( ) maps 3 to 0 since (3) = 3 = 9 and (9) = 0 I and are described by an equation then oten ( ) can be described by an equation Here ( )(x) = ((x)) = (x ) = x + So ( )(x) = x + 7 / 0 On the other hand ( )(x) = ((x)) = (x + ) = (x + ) So ( )(x) = x + but ( )(x) = (x + ) 8 / 0

3 Function composition examples (x) =, (x) = x + In elementary alebra, we learned the importance o parentheses that + is quite dierent rom ( + x) Part, Functions Lecture 5b, : Examples The use o parentheses and the order o operations is especially important in the composition o unctions Here squarin and then addin one ( ) is dierent rom addin one and then squarin ( ) Dr Ken W Smith ( )(x) = + but ( )(x) = (x + ) In the next lesson we will work some problems involvin unction composition 9 / 0 Exercises ( )(x) ( )(x) 0 / 0 Exercises Some worked examples Given the unctions (x) = and (x) = x +, create the ollowin composition unctions: Given the unctions and, below, ind the composition unctions and The unction ( )(x) is the same as ((x)); ( )(x) is the same as ( (x)) (x) = + and (x) = 3 Solution Solutions ( )(x) = ((x)) = (x + ) = (x + ) = + 4x + 4 = + 4x + 3 ( )(x) = ( (x)) = ( ) = ( ) + = + / 0 (x) = + and (x) = 3 ( )(x) = ((x)) = ( 3) = 3 + = 3 + = 4 ( )(x) = ( (x)) But (anythin) = 3, so the answer is 3 ( )(x) = 4 and ( )(x) = 3 / 0

4 Exercises 3 (x) = + 9 and (x) = 4 (x) = + 5 and (x) = Function Chainin x x 5 Solutions 3 4 (x) = + 9 and (x) = x ( )(x) = ((x)) = ( x) = ( x) + 9 = x + 9 ( )(x) = ( (x)) = (x + 9) = + 9 ( )(x) = x + 9 and ( )(x) = + 9 Next we will look at chainin unctions toether with unction composition (x) = + 5 and (x) = x 5 ( )(x) = ((x)) = ( x 5) = ( x 5) + 5 = (x 5) + 5 = x ( )(x) = ( (x)) = (x + 5) = x = = x ( )(x) = x and ( )(x) = x 3 / 0 4 / 0 Breakin a unction down into components It is convenient at times to break a unction down into pieces, so that we may view the unction itsel as a composition o two or more unctions For example, suppose h(x) = 3x + 4 Part, Functions Lecture 5c, More I we input an x-value into h, we irst compute 3x + 4 and then we take the square root So we may view the unction h as a composition o a unction (x) = 3x + 4 and (x) = x Dr Ken W Smith h = ( ) where (x) = 3x + 4 and (x) = x This is an example o takin a complicated unction and breakin it down into its simple pieces 5 / 0 6 / 0

5 Function Chainin Breakin a unction down Some more worked examples For each o the unctions (x) and h(x) below, ind a unction (x) such that h(x) = ( )(x) (x) = 0 x, h(x) = 0 ( 7) (x) = x, h(x) = x + 4 Solution h(x) = 0 ( 7) = ( )(x) i (x) = x 7 h(x) = x + 4 = ( )(x) i (x) = x + 4 For each unction h iven below, decompose h into the composition o two unctions and so that h = h(x) = (x + 5) h(x) = 3 5x + 3 h(x) = cos x Solutions h(x) = (x + 5) is the composition o (x) = x + 5 and (x) = x h(x) = 3 5x + is the composition o (x) = 5x + and (x) = 3 x 3 h(x) = cos x is the composition o (x) = cos x and (x) = x (We can ind the unctions and, even i we have not yet studied the unction cos x the notation leads us to the answer!) 7 / 0 Function Chainin 8 / 0 Function Chainin There is no limit to the number o unctions we can chain toether! Once we understand unction composition, there is no reason to stop at composin just two unctions! We can compose a chain o unctions, runnin an input x throuh one unction ater another For example, suppose that (x) = x, (x) = 3x + 5 and h(x) = x I we run x throuh, and h in that order we et (h )(x) = h(((x))) = h((x )) = h(3x + 5) = 3x / 0 For example, suppose that (x) = x, (x) = 3x + 5, h(x) = x and j(x) = cos(x) I we run x throuh,, h and j in that order we et (j h )(x) = j(h(((x)))) = j(h((x ))) = j(h(3x + 5)) = j( 3x + 5) = cos( 3x + 5) (We can do this even i we have not yet studied the cosine unction cos(x) we just ollow our notation!) In calculus, ater we study the derivative o a unction, we will learn to take the derivative o a chain o unctions composed toether in this manner The method we develop there is called The Chain Rule or derivatives 0 / 0