HW03 - f (-2) = (-2 + 1) 3 =-1, m = f (-2) = 3(-2) 2 +...

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MATH 132 HW 03 Answer Section 3.1 [2]. Find the derivative of F ( x ) = ( x - 1) 2 + 1 Solution: F ( x + h ) - F ( x ) h = [( x + h - 1) 2 + 1] - [( x - 1) 2 + 1] h = ( x + h - 1) 2 + 1 - ( x - 1) 2 - 1 h = ( x + h - 1) 2 - ( x - 1) 2 h = [( x + h - 1) - ( x - 1)] · [( x + h - 1) + ( x - 1)] h = [ h ] · [2 x + h - 2] h = 2 x + h - 2 2 x - 2 as h 0 = F 0 ( x ) = 2 x - 2 = F 0 ( - 1) = - 4, F 0 (0) = - 2, F 0 (2) = 2 Way 2: F ( x + h ) - F ( x ) h = [( x + h - 1) 2 + 1] - [( x - 1) 2 + 1] h = [( x - 1) + h ] 2 + 1 - ( x - 1) 2 - 1 h = ( x - 1) 2 + 2( x - 1) h + h 2 - ( x - 1) 2 h = 2( x - 1) h + h 2 h = 2( x - 1) + h 2( x - 1) as h 0 [4]. Find the derivative of k ( z ) = 1 - z 2 z Solution: k ( z + h ) - k ( z ) h = 1 - ( z + h ) 2( z + h ) - 1 - z 2 z h = 1 h · (1 - z - h ) z - ( z + h )(1 - z ) 2( z + h ) z = 1 h · z - z 2 - hz - ( z - z 2 + h - hz ) 2( z + h ) z = 1 h · z - z 2 - hz - z + z 2 - h + hz 2( z + h ) z = 1 h · - h 2( z + h ) z = - 1 2( z + h ) z - 1 2 z 2 as h 0 = k 0 ( x ) = - 1 2 z 2 = k 0 ( - 1) = - 1 2 , k 0 (1) = - 1 2 , k 0 ( 2) = - 1 4 1
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[16]. Differentiate y = ( x + 1) 3 and the equation of the tangent line at x = - 2 Solution: Let f ( x ) = ( x + 1) 3 f ( x + h ) - f ( x ) h = ( x + h + 1) 3 - ( x + 1) 3 h = ( x 3 + 3 hx 2 + 3 x 2 + 3 h 2 x + 6 hx + 3 x + h 3 + 3 h 2 + 3 h + 1) - ( x 3 + 3 x 2 + 3 x + 1) h = x 3 + 3 hx 2 + 3 x 2 + 3 h 2 x + 6 hx + 3 x + h 3 + 3 h 2 + 3 h + 1 - x 3 - 3 x 2 - 3 x - 1 h = 3 hx 2 + 3 h 2 x + 6 hx + h 3 + 3 h 2 + 3 h h = h (3 x 2 + 3 hx + 6 x + h 2 + 3 h + 3) h = 3 x 2 + 3 hx + 6 x + h 2 + 3 h + 3 3 x 2 + 6 x + 3 as h 0 = y 0 = f 0 ( x ) = 3 x 2 + 6 x + 3 =
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Unformatted text preview: f (-2) = (-2 + 1) 3 =-1, m = f (-2) = 3(-2) 2 + 6(-2) + 3 = 3 = the equation of the tangent line is y-(-1) = 3( x-(-2)) = y + 1 = 3( x + 2) Way 2: f ( x + h )-f ( x ) h = ( x + h + 1) 3-( x + 1) 3 h = [( x + 1) + h ] 3-( x + 1) 3 h = ( x + 1) 3 + 3( x + 1) 2 h + 3( x + 1) h 2 + h 3-( x + 1) 3 h = 3( x + 1) 2 h + 3( x + 1) h 2 + h 3 h = [3( x + 1) 2 + 3( x + 1) h + h 2 ] h h = 3( x + 1) 2 + 3( x + 1) h + h 2 3( x + 1) 2 as h 2...
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HW03 - f (-2) = (-2 + 1) 3 =-1, m = f (-2) = 3(-2) 2 +...

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