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

Algebra/Trig ReviewFirst, let’s recall that for b > 0 and b ≠ 1 an exponential function is any function that is inthe formxf ( x) = bWe require b ≠ 1 to avoid the following situation,xf x = =( ) 1 1So, if we allowed b = 1 we would just get the constant function, 1.We require b > 0 to avoid the following situation,x1f ( x) = ( −4) ⇒ f ⎛ ⎞ ( 4) 1 2⎜ ⎟= − = −4⎝2⎠By requiring b > 0 we don’t have to worry about the possibility of square roots ofnegative numbers.1. Evaluate f ( x ) = 4 x , g( x)x⎛1⎞−= ⎜ ⎟ and h( x) = 4 x at x =−2, − 1, 0,1, 2 .⎝4⎠SolutionThe point here is mostly to make sure you can evaluate these kinds of functions. So,here’s a quick table with the answers.x = − 2 x = − 1 x = 0 x = 1 x = 2211f − 1 =16 4f ( 0)= 1 f ( 1)= 4 f ( 2)= 162 16 g − 1 = 4 g ( 0)= 11 1g ( 1)= g ( 2)=4 162 16 h − 1 = 4 h ( 0)= 11 1h ( 1)= h ( 2)=4 16f ( x ) f ( − ) = ( )g( x ) g ( − ) = ( )h( x ) h( − ) = ( )Notice that the last two rows give exactly the same answer. If you think about it thatshould make sense because,x x⎛1⎞1 1 − xg( x) = ⎜ ⎟ = = = 4 = hx x( x)⎝4⎠4 4x⎛1⎞−f x = , g( x)= ⎜ ⎟ and h( x) = 4 x on the same axis system.⎝4⎠2. Sketch the graph of ( ) 4 xSolutionNote that we only really need to graph f ( x ) and ( )previous Problem that g( x) h( x)g x since we showed in the= . Note as well that there really isn’t too much todo here. We found a set of values in Problem 1 so all we need to do is plot the pointsand then sketch the graph. Here is the sketch,© 2006 Paul Dawkins 82http://tutorial.math.lamar.edu/terms.aspx