241ex2e - Calculus I (Math 241) Test 2 (No Work No Credit)...

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Unformatted text preview: Calculus I (Math 241) Test 2 (No Work No Credit) Problem 1. [35 Points] Consider the function f ( x ) = x x 2 + 1 . Discuss the graph of f . Justify your answers. 1. Find the intercepts and decide where the function is positive and where it is negative. Decide on symmetry properties and find the asymptotes. 2. Find the critical points, the intervals on which the function is increas- ing, and the intervals on which it is decreasing. 3. Find the intervals on which f is concave up or concave down, respec- tively. 4. Find the local minima, maxima and the inflection points of f . 5. Sketch the graph of f based on your computation. On the graph, indicate the local extrema and the inflection points. 6. Extra Credit [5 Points] Give a proper argument why the local extrema turn out to be the absolute ones. Solution: 1.) Clearly, f ( x ) = 0 if and only if x = 0, so that the graph intersects the coordinate axes in (0 , 0), and only there. The denominator x 2 + 1 of the expression for f ( x ) is always positive, so the sign of f ( x ) depends only on the numerator, which is x . We deduce that f ( x ) is positive when x > 0, and that f ( x ) is negative when x < 0. The function is odd ( f (- x ) =- f ( x )), so that the graph is symmetric with respect to the origin of the coordinate system, (0 , 0). The x axis ( y = 0) is the horizontal asumptote of the function. 2.) We use the quotient rule to calculate: f ( x ) = ( x 2 + 1)- x 2 x ( x 2 + 1) 2 = 1- x 2 ( x 2 + 1) 2 = (1 + x )(1- x ) ( x 2 + 1) 2 ....
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This note was uploaded on 07/26/2009 for the course MATH 215 taught by Professor Heiner during the Spring '09 term at Hawaii.

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241ex2e - Calculus I (Math 241) Test 2 (No Work No Credit)...

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