{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

131f11x1ANS - Math 131 Exam 1 Solutions 1(a[8 The average...

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

View Full Document Right Arrow Icon
Math 131 Exam 1 Solutions 10/6/11 1. (a) [8%] The average rate of change is: P (200) - P (100) 200 - 100 = 2000 / 200 - 2000 / 100 100 = 10 - 20 100 = - 1 10 lb / in 2 in 3 (b) [8%] The instantaneous rate of change is the limit: lim h 0 P (100 + h ) - P (100) h = lim h 0 2000 100 + h - 2000 100 h = 2000 lim h 0 1 100 + h - 1 100 h = 2000 lim h 0 100 - (100 + h ) h 100 (100 + h ) = 2000 lim h 0 - h h 100 (100 + h ) = 2000 lim h 0 - 1 100 (100 + h ) = 2000 - 1 100(100 + 0) = - 1 5 lb / in 2 in 3 2. [16%] f 0 ( x ) = lim h 0 f ( x + h ) - f ( x ) h = lim h 0 [( x + h ) 2 - ( x + h ) + 5] - [ x 2 - x + 5] h = lim h 0 [( x 2 + 2 x h + h 2 ) - x - h + 5] - [ x 2 - x + 5] h = lim h 0 x 2 + 2 x h + h 2 - x - h + 5 - x 2 + x - 5 h = lim h 0 2 x h + h 2 - h h = lim h 0 h (2 x + h - 1) h = lim h 0 (2 x + h - 1) = 2 x + 0 - 1 [by Direct Substitution] = 2 x - 1 1
Background image of page 1

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

View Full Document Right Arrow Icon
3. [16%] There are many possibilities. Here is one: Minus 4 Minus 3 Minus 2 Minus 1 1 2 3 4 x Minus 4 Minus 3 Minus 2 Minus 1 1 2 3 4 y 2
Background image of page 2
4. (a) [5%] lim x → - 3 x 2 - x - 12 x + 3 = lim x → - 3 ( x + 3) ( x - 4) x + 3 [factor top and bottom] = lim x → - 3 ( x - 4) [cancel common factor] = - 3 - 4 [Direct Substitution] = - 7 (b) [5%] lim x 1 x 2 + 3 x + 5 x 2 - 4 = (1) 2 + 3(1) + 5 (1) 2 - 4 [Direct Substitution] = 9 - 3 = - 3 (c) [5%] Since | cos x | ≤ 1 for all x , then for all x , e x cos x ≤ | e x | = e x .
Background image of page 3

Info iconThis 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 ]}