test2pbcalculusIfall06sol - Roll No x 1 Value Theorem for y...

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Roll No Name: Test # 2, Solutions Math 1131 1 . Let f ( x ) = x x 2 + 1 defined on the interval [0 , 1] . Check the hypotheses of the Intermediate Value Theorem for y = 2 / 5 . Find the value c in the interval [0 , 1] such that f ( c ) = 2 / 5 . Solution: The function f is a rational function defined everywhere and so it is continuous at every point in its domain of definition as all the elementary functions. Because f (0) = 0 and f (1) = 1 2 we need to check that y = 2 / 5 is in the interval [0 , 1 / 2]. This is clearly true and so the Intermediate Value Theorem applies in this cases: the equation f ( x ) = y must have at least a solution in the interval (0 , 1). In fact f ( x ) = 2 5 is equivalent to 5 x = 2 x 2 + 2 or 2 x 2 - 5 x +2 = 0. Factoring this quadratic will give (2 x - 1)( x - 2) = 0 which has solutions x 1 = 2 and x 2 = 1 / 2. 2 . For the function in Problem 1 find the equation of the tangent line to the graph of y = f ( x ) at the point (3 , 3 / 10) . Use a graphing calculator to draw the graph of y = f ( x ) and the above tangent line. Solution: If we calculate the derivative of this function we obtain f 0 ( x ) = x 2 + 1 - x (2 x ) ( x 2 + 1) 2 = 1 - x 2 ( x 2 + 1) 2 = f 0 (3) = - 8 100 = - 2 25 and then the equation of the tangent line is y = f (3) + f 0 (3)( x - 3) ⇐⇒ y = 3 10 - 2 25 ( x - 3) ⇐⇒ y = 27 - 4 x 50 , and its graph together with the graph of y = f ( x ): 3 2 1 0 0.4 x 0.2 0.3 5 0 4 0.5 0.1 3 . The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t(measured in seconds) is Q ( t ) = 17 t 2 - 4 t 3 - 24 t . Find at what time is the current at a maximum?
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