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Questions:



 1. If f  x  = 〚 x〛 x then what is f '  x at x = a∈ℤ ?

 2. If ∣ f  x −1∣  x 2 e x for all x ∈ℝ then lim f  x as x 0 is ?

                                                                   sinx
 3. Use the Intermediate Value Theorem to show that f  x =            has at least one root on the interval [2,5]
                                                                  x 5−7



Solutions:


 Solution 1:
        We use the definition of the derivative.
        CASE 1: as x  a
                     f  x − f a        〚 x〛x−〚a〛−a         x −a
                lim                  = lim                = lim       =1
                          x−a                  x−a              x −a
        CASE 2 :as x  a−
                     f  x − f a        〚 x〛x−〚a〛−a      a−1a−a−a
                lim                  = lim                              =∞
                          x−a                  x−a                 0-
        Thus , the derivative of f does not exist at x=a ∈ℤ

 Solution 2 :
        We use the SqueezeTheorem.
        ∣ f  x −1∣  x 2 e x for all x∈ℝ
        - x 2 e x  f  x −1x 2 e x for all x∈ℝ
        as x  0, lim - x 2 e x   lim  f  x−1  lim  x 2 e x  for all x∈ℝ
        as x  0, lim  f  x −1 = 0 for all x∈ℝ
        as x  0, lim f  x = 1 for all x ∈ℝ

 Solution 3:
        Note that onthe interval 0,  , sinx0
        Since 2∈[0,] , f 20
        Note that onthe interval  , 2  , sinx0
        Since5∈ , 2 , f 50
        Thus ,it follows from the Intermediate Value Theorem that
        ∃ c∈[2,5] such that f  c=0, ie. f has a zero on the interval [2,5]

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Math 53 sample questions from jamie pdf

  • 1. Questions: 1. If f  x  = 〚 x〛 x then what is f '  x at x = a∈ℤ ? 2. If ∣ f  x −1∣  x 2 e x for all x ∈ℝ then lim f  x as x 0 is ? sinx 3. Use the Intermediate Value Theorem to show that f  x = has at least one root on the interval [2,5] x 5−7 Solutions: Solution 1: We use the definition of the derivative. CASE 1: as x  a f  x − f a  〚 x〛x−〚a〛−a x −a lim = lim = lim =1 x−a x−a x −a CASE 2 :as x  a− f  x − f a  〚 x〛x−〚a〛−a a−1a−a−a lim = lim  =∞ x−a x−a 0- Thus , the derivative of f does not exist at x=a ∈ℤ Solution 2 : We use the SqueezeTheorem. ∣ f  x −1∣  x 2 e x for all x∈ℝ - x 2 e x  f  x −1x 2 e x for all x∈ℝ as x  0, lim - x 2 e x   lim  f  x−1  lim  x 2 e x  for all x∈ℝ as x  0, lim  f  x −1 = 0 for all x∈ℝ as x  0, lim f  x = 1 for all x ∈ℝ Solution 3: Note that onthe interval 0,  , sinx0 Since 2∈[0,] , f 20 Note that onthe interval  , 2  , sinx0 Since5∈ , 2 , f 50 Thus ,it follows from the Intermediate Value Theorem that ∃ c∈[2,5] such that f  c=0, ie. f has a zero on the interval [2,5]