Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewDo not let the fact that there are x’s in the parenthesis on the left get you workedup! Simply replace all the x’s in the formula on the right side with x − 3 . Thisone works exactly the same as(f).(h) ( ) ( ) ( )2 2f 4x− 1 =− 4x− 1 + 6 4x−1 − 11 =− 16x + 32x−18Do not get excited by problems like (e) – (h). This type of problem works the same as(a) – (d) we just aren’t using numbers! Instead of substituting numbers you aresubstituting letters and/or other functions. So, if you can do (a) – (d) you can dothese more complex function evaluations as well!2. Given f ( x ) = 10 find each of the following.(a) f ( 7)(b) f ( 0)(c) f ( − 14)SolutionThis is one of the simplest functions in the world to evaluate, but for some reasonseems to cause no end of difficulty for students. Recall from the previous problemhow function evaluation works. We replace every x on the right side with what everis in the parenthesis on the left. However, in this case since there are no x’s on theright side (this is probably what causes the problems) we simply get 10 out of each ofthe function evaluations. This kind of function is called a constant function. Just tobe clear here are the function evaluations.( ) ( ) ( )f 7 = 10 f 0 = 10 f − 14 = 1023. Given f ( x) = 3x − x+ 10 and g( x) 1 20⎛ f ⎞⎜ ⎟⎝ g ⎠f g x(a) ( f − g)( x)(b) ( x)= − x find each of the following.(c) ( fg )( x )(d) ( f g)( 5)(e) ( )( ) (f) ( gf )( x)SolutionThis problem makes sure you are familiar with notation commonly used withfunctions. The appropriate formulas are included in the answer to each part.(a)( f − g)( x) = f ( x) −g( x)2= 3x − x+ 10 −( 1−20x)2= 3x − x+ 10 − 1+20x2= 3x+ 19x+9(b)© 2006 Paul Dawkins 14http://tutorial.math.lamar.edu/terms.aspx