2010 Calculus Test 1 - 1 Math 1151 Calculus S1 2010 Test 1...

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Unformatted text preview: 1. Math 1151 Calculus S1 2010 Test 1 Version 2a (1) On the same axes sketch the graphs of y = | 3 x- 1 | and y = x + 4, and hence or otherwise solve | 3 x- 1 | = x + 4. Solution. From the graph, there are two solutions, given by 3 x- 1 = x + 4 and 3 x- 1 =- x- 4. So x = 5 / 2 ,- 3 / 4. (2) Show that f ( x ) = x 3 + 6 x 2 + 12 x + 7 considered as a function on R has an inverse g ( x ). Since f ( x ) = 3( x + 2) 2 > 0, the function f is increasing on its domain, hence is one-to-one. It is surjective since any cubic polynomial has at least one real root. This shows f has an inverse. (3) Use the fact that d d x tanh x = sech 2 x to find d y d x when y = tanh- 1 3 x . Solution. Let f ( x ) = tanh- 1 x . Then d f d x ( x ) = 1 d d x tanh( f ( x )) = 1 sech( f ( x )) 2 = 1 1- tanh( f ( x )) 2 = 1 1- x 2 d y d x ( x ) = 3 1- 9 x 2 . (4) Determine the limiting behaviour of f ( x ) = 1 + 2 x 2 + sin x x 3 + cos x as x → ∞ ....
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2010 Calculus Test 1 - 1 Math 1151 Calculus S1 2010 Test 1...

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