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Unformatted text preview: saliyev (is4663) – Homework 3 – knopf – (55420) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This assignment covers Sections 3.3, 3.4, and 3.5 001 10.0 points Find the value of f ′ ( a ) when f ( t ) = 2 t + 7 t + 6 . 1. f ′ ( a ) = − 5 ( a + 6) 2 2. f ′ ( a ) = 5 ( a + 6) 2 correct 3. f ′ ( a ) = − 6 a + 6 4. f ′ ( a ) = 6 ( a + 6) 2 5. f ′ ( a ) = 5 a + 6 6. f ′ ( a ) = − 6 ( a + 6) 2 Explanation: By definition, f ′ ( a ) = lim h → f ( a + h ) − f ( a ) h . Now, for the given f , f ( a + h ) = 2( a + h ) + 7 a + h + 6 , while f ( a ) = 2 a + 7 a + 6 . Thus f ( a + h ) − f ( a ) = 2( a + h ) + 7 a + h + 6 − 2 a + 7 a + 6 = parenleftBigg (2( a + h ) + 7)( a + 6) − ( a + h + 6)(2 a + 7) parenrightBigg ( a + h + 6)( a + 6) . But { 2( a + h ) + 7 } ( a + 6) = 2 h ( a + 6) + (2 a + 7)( a + 6) , and ( a + h +6)(2 a +7) = h (2 a +7)+( a +6)(2 a +7) . Consequently, f ( a + h ) − f ( a ) h = h { 2( a + 6) − (2 a + 7) } h ( a + h + 6)( a + 6) = 5 ( a + h + 6)( a + 6) , in which case f ′ ( a ) = 5 ( a + 6) 2 . 002 10.0 points Find the xcoordinate of all points on the graph of f ( x ) = x 3 − 2 x 2 − 4 x + 5 at which the tangent line is horizontal. 1. xcoord = 2 3 2. xcoords = 2 3 , − 2 3. xcoord = 2 4. xcoord = − 2 5. xcoord = − 2 3 6. xcoords = − 2 3 , 2 correct Explanation: saliyev (is4663) – Homework 3 – knopf – (55420) 2 The tangent line will be horizontal at P ( x , f ( x )) when f ′ ( x ) = 0 . Now f ′ ( x ) = 3 x 2 − 4 x − 4 = (3 x + 2)( x − 2) . Consequently, x = − 2 3 , 2 . 003 10.0 points Determine the derivative of f when f ( x ) = √ 6 x + 1 √ 6 x . 1. f ′ ( x ) = 1 2 parenleftbigg 6 x − 1 √ 6 x parenrightbigg 2. f ′ ( x ) = 6 x − 1 x √ 6 x 3. f ′ ( x ) = 6 x + 1 x √ 6 x 4. f ′ ( x ) = 1 2 parenleftbigg 6 x + 1 x √ 6 x parenrightbigg 5. f ′ ( x ) = 6 x + 1 √ 6 x 6. f ′ ( x ) = 1 2 parenleftbigg 6 x − 1 x √ 6 x parenrightbigg correct Explanation: Since d dx a x r = r a x r − 1 for all exponents r negationslash = 0 and constants a , we see that f ′ ( x ) = 1 2 radicalbigg 6 x − 1 2 1 x √ 6 x . After simplification this becomes f ′ ( x ) = 1 2 parenleftbigg 6 x − 1 x √ 6 x parenrightbigg . 004 10.0 points Find the derivative of g ( x ) = parenleftbigg x + 2 x + 1 parenrightbigg (2 x − 5) . 1. g ′ ( x ) = 2 x 2 + 4 x + 9 x + 1 2. g ′ ( x ) = x 2 + 4 x − 9 ( x + 1) 2 3. g ′ ( x ) = 2 x 2 − 4 x − 9 x + 1 4. g ′ ( x ) = 2 x 2 − 4 x − 9 ( x + 1) 2 5. g ′ ( x ) = 2 x 2 + 4 x + 9 ( x + 1) 2 correct 6. g ′ ( x ) = x 2 − 4 x + 9 x + 1 Explanation: By the Quotient and Product Rules we see that g ′ ( x ) = 2 braceleftbigg x + 2 x + 1 bracerightbigg + (2 x − 5) braceleftbigg ( x + 1) − ( x + 2) ( x + 1) 2 bracerightbigg = 2 braceleftbigg x + 2 x + 1 bracerightbigg + braceleftbigg 2 x − 5 ( x + 1) 2 bracerightbigg...
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This note was uploaded on 02/28/2011 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam
 Calculus

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