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Unformatted text preview: Name: 1 1. Find the domains of the following functions: (a) f ( x ) = log( x ( x + 1)) Logs only take positive values, so th domain will be x values such that x ( x + 1) > 0. The critical points for this inequality are x = 0 and x = 1. Moreover, the graph forms an upward facing parabola, so its positive values are those smaller than 1 or larger than 0. Hence Dom( f ) = ( , 1) (0 , ). (b) g ( x ) = x 3 3 x 2 + 4 x + 17 This is a polynomial; no zero divisions, roots, logs, or context, so it is everywhere defined, i.e. Dom( g ) = ( , ). (c) h ( x ) =  2 x 2 + 4 x 1 Here we have a square root, so we need the inside to be nonneg ative (note zeroes ARE allowed: 0 = 0). So we test 2 x 2 + 4 x 1 0; using the quad formula we get x = 1 2 for the roots. This parabola faces downwards, so the positive bit is the middle interval. Thus the domain for our square root function is Dom( h ) = [ 1 2 , 1 + 2]. (d) k ( x ) = x/ ( x + 1) Here we have two possible problems: a square root and a zero division. The zero division occurs at x = 1, so we know that point isnt in the domain. The x part of the function implies we can only input nonnegative numbers anyway though, so our function is defined on [0 , ). 2. Calculate the difference quotient [ f ( x + h ) f ( x )] /h of the following functions: (a) f ( x ) = x + 1 Compute f ( x + h ) = ( x + h ) + 1 = x h + 1. Then f ( x + h ) f ( x ) h = x h + 1 ( x + 1) h = h h = 1 . Name: 2 (b) f ( x ) = x x 2 Compute f ( x + h ) = ( x + h ) ( x + h ) 2 = x + h x 2 2 xh h 2 . Then f ( x + h ) f ( x ) h = x + h x 2 2 xh h 2 ( x x 2 ) h = h 2 xh h 2 h = 1 2 x h (c) f ( x ) = 1 2 x Compute f ( x + h ) = 1 2( x + h ) = 1 2 x +2 h . Note that f ( x + h ) f ( x ) = 1 2 x + 2 h 1 2 x = x 2 x ( x + h ) x + h x (2 x + 2 h ) = h x (2 x + 2 h ) . Then we have f ( x + h ) f ( x ) h = h xh (2 x + 2 h ) = 1 x (2 x + 2 h ) 3. Use the definition of the derivative to evaluate f ( x ) for: (a) f ( x ) = 2 x 7 As above, we compute the difference quotient and take the limit as h 0. f ( x + h ) = 2( x + 5) 7 = 2 x + 2 h 7, so f ( x + h ) f ( x ) h = 2 x + 2 h 7 (2 x 7) h = 2 h/h = 2 . The limit as x a of a constant function is constant for any value a , so f ( x ) = 2. (b) f ( x ) =  x  f ( X ) =  x  = x x  x x < . Then the right difference quotient (values to the right of 0) is f ( x + h ) f ( x ) h = x + h x h = h/h = 1 Name: 3 while the left difference quotient (values left of zero) is f ( x + h ) f ( x ) h = x h  x h = h/h = 1 ....
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This note was uploaded on 04/16/2008 for the course CALC 1304 taught by Professor Pruett during the Spring '08 term at Baylor.
 Spring '08
 pruett
 Critical Point

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