Test 2 practice problems-solutions

Test 2 practice problems-solutions - Practice Problems for...

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Practice Problems for Test 2 Answers 1) Use the definition: ¢ f ( x ) = lim h 0 f ( x + h ) - f ( x ) h to show that ¢ f ( x ) = 4 x - 3 if f ( x ) = 2 x 2 - 3 x f ( x ) = 2 x 2 - 3 x f ( x + h ) = 2( x + h ) 2 - 3( x + h ) = 2( x 2 + 2 xh + h 2 ) - 3( x + h ) ¢ f ( x ) = lim h 0 f ( x + h ) - f ( x ) h ¢ f ( x ) = lim h 0 (2 x 2 + 4 xh + 2 h 2 - 3 x - 3 h ) - (2 x 2 - 3 x ) h lim h 0 h (4 x + 2 h - 3) h = 4 x + 2(0) - 3 = 4 x - 3 2) Use g ( x ) = x + 1 if x £ 2 2 x - 1 if x > 2 a) Is g ( x ) continuous for all x ? Each "piece" is continuous so we just need to check where they meet: lim x 2 - x + 1 = 2 + 1 = 3 lim x 2 + 2 x - 1 = 2(2) - 1 = 3 g (2) = 2 + 1 = 3 Therefore g ( x ) is continuous. b) Is g ( x ) differentiable at for all x ? Each "piece" is differentiable so we just need to check where they meet: ¢ g ( x ) = 1 if x < 2 2 if x > 2 Therefore g ( x ) is NOT differentiable at x = 2 . ¢ g - (2) ¢ g + (2) . c) Sketch the graphs of g ( x ) and ¢ g ( x ) .
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3) Find dy dt for each of the following: a) dy dt = - 9 t 5 cos 2 ( t 3 )sin( t 3 ) + 3 t 2 cos 3 ( t 3 ) b) dy dt = 1 - 5sin(5 t )cos(5 t ) - 3cos(3 y )sin(3 y ) c) dy dt = 5 10 t - 5 d) dy dt = ( t 2 + 5) 9 ( - 21 t 2 + 20 t - 5) e) dy dt = 2 p 2 t cos( p 2 t 2 ) f) dy dt = t 2 [ t sec 2 ( t - 1) + 3tan( t - 1)] g) dy dt = ( - sin t )(cos(cos t )) h) dy dt = 0 i) dy dt = - sin t 1 + cos 2 t j) dy dt = e 5 t 5(1 + 3 t ) - ln3(3
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