{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

practice_test_1_key

practice_test_1_key - Name 1 Find the domains of the...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 isn’t 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 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 0 ( 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 0 ( x ) = 2. (b) f ( x ) = | x | f ( X ) = | x | = x x 0 - x x < 0 . 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
Image of page 2
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 . So the left limit is - 1 while the right hand limit is 1, so the limits don’t agree. But the derivative is defined to be THE limit of the AROC functions as h tends to zero, and this limit does not exist.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}