Exam1_S2010 Solutions

# Sin 2 x careful this is not a pythagorean identity

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sin 2 ( x ) . (Careful, this is not a Pythagorean identity!) Thus we must solve the equation cos 2 ( x ) - sin 2 ( x ) = 0, or cos 2 ( x ) = sin 2 ( x ) . Because of the squares, the equation is true when cos( x ) = ± sin( x ). In the interval (0 , π ), the solutions are x = π/ 4 and x = 3 π/ 4. (continued on next page...) 3

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This gives us the sign chart 0 π 4 3 π 4 π Between each two points, we choose values and check the sign of f 0 ( x ). For example, f 0 ( π/ 6) = 3 2 ! 2 - 1 2 2 = 3 4 - 1 4 > 0 Similarly we can find that f 0 ( π/ 2) < 0 and f 0 (5 π/ 6) > 0. Thus our sign chart is (+) ( - ) (+) 0 π 4 3 π 4 π Thus f has a maximum at x = π/ 4 and a minimum at x = 3 π/ 4. 8. Let f ( x ) = ( x - 2) 2 + 6 and g ( x ) = 3 x . a. Graph both functions. Observe that f ( x ) is the parabola x 2 shifted up 6 units and to the right 2 units. b. Determine (algebraically) the points of intersection. Two graphs of functions f ( x ) and g ( x ) intersect when f ( x ) = g ( x ). Thus we solve the equation ( x - 2) 2 + 6 = 3 x. Expanding the square and rearranging, we get x 2 - 7 x + 10 = 0, which factors as ( x - 5)( x - 2) = 0. Thus the points of intersection occur at x = 2 and x = 5. c. What is the area of the region bounded by the two curves? Observe that on the interval [2 , 5], the function g is larger than f . Thus we compute the integral Z 5 2 3 x - ( x - 2) 2 - 6 dx = Z 5 2 7 x - x 2 - 10 dx = 7 2 x 2 - 1 3 x 3 - 10 x 5 2 = 7 2 (5) 2 - 1 3 (5) 3 - 10(5) - 7 2 (2) 2 + 1 3 (2) 3 + 10(2) . 4
8. Suppose the rate at which a population of pumas is increasing is given by R ( t ) = 500 te 0 . 08 t measured in pumas/year. If the initial population of pumas is 1000, estimate the population after 6 years.

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