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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 onetoone. 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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 Three '11
 Calculus, lim, dx

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