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Sullivan  04 apcalc4e 45342 ch02 166 233 5pp  August 7, 2023  12:54



                                                                         Section 2.2 • The Derivative as a Function; Differentiability  183

                                                                    y                                    y
                                  y                                 2                                    2
                                  2
                                                                 (0, 0)
                                                        4    2          2     4  x         24    22      (0, 0)  2  4  x
                                                                                                 f has a cusp
                            2           2   x                        f has a vertical
                                                                   2                             at (0, 0)  22  f has a vertical
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                                                                      tangent line at (0, 0)               tangent line at (0, 0)
                                     f has a corner
                                 2
                                     at (0, 0)                     x   0                                x 5 0
                                                                    √                                     2/3
                                                                    3
                   Figure 13 f (x) = |x|;            Figure 14 f (x) =  x;                Figure 15 f (x) = x  ;
                            f (0) does not exist.             f (0) does not exist.                f (0) does not exist.
                                                                                                    ′
                             ′
                                                               ′
                                                      EXAMPLE 6 Identifying Where a Function Is Not Differentiable
                                                                                                2
                                                                                            −2x + 4   if x < 1
                                                      Given the piecewise defined function f (x) =  2        ,
                                                                                               x + 1  if x ≥ 1
                                                      determine whether f (1) exists.
                                                                       ′
                                                      Solution
                                                      Use Form (1) of the definition of a derivative to determine whether f (1) exists.
                                                                                                             ′
                                                                       f (x) − f (1)   f (x) − 2
                                                                                                        2
                                                                   lim           = lim            f (1) = 1 + 1 = 2
                                                                   x→1   x − 1     x→1  x − 1
                                                                                                      2
                                                                              2
                                                         If x < 1, then f (x) = −2x + 4; if x ≥ 1, then f (x) = x + 1. So, it is necessary to
                                                      find the one-sided limits at 1.
                                                                                                     2
                                                                                  2
                                                             f (x) − f (1)   (−2x + 4) − 2       −2(x − 1)
                                                         lim           = lim              = lim
                            y      2
                               y   x    1               x→1 −  x − 1     x→1 −    x − 1     x→1 −  x − 1
                            6
                                                                                (x − 1)(x + 1)
                            5                                          = −2 lim              = −2 lim (x + 1) = −4
                                        Slope   2                           x→1 −   x − 1        x→1 −
                            4                                                  2
                                                             f (x) − f (1)   (x + 1) − 2      (x − 1)(x + 1)
                                                         lim           = lim            = lim             = lim (x + 1) = 2
                            3                           x→1 +  x − 1     x→1 +  x − 1     x→1 +   x − 1     x→1 +
                      2
                y    2x    4
                              (1, 2)
                            2        f  has a corner at (1, 2)                                   f (x) − f (1)
                                     f (1) does not exist  Since the one-sided limits are not equal, lim   does not exist, and
                                                                                             x→1   x − 1
                            1                         so f (1) does not exist.
                                                         ′
                                    Slope    4
                                                         Figure 16 illustrates the graph of the function f from Example 6. At 1, where the
                     2   1       1   2    3   4  x
                                                      derivative does not exist, the graph of f has a corner. We usually say that the graph of f
                  Figure 16 f has a corner at (1, 2).  is not smooth at a corner.
                                                                                R
                                                      NOW WORK    Problem 39 and AP Practice Problems 1, 5, 9, and 10.
                                                         Example 7 illustrates the behavior of the graph of a function f when the derivative
                                                                                       f (x) − f (c)
                                                      at a number c does not exist because lim    is infinite.
                                                                                   x→c    x − c
                                                      EXAMPLE 7 Showing That a Function Is Not Differentiable
                                                  CALC CLIP
                                                      Show that f (x) = (x − 2) 4/5  is not differentiable at 2.
                                                      Solution
                                                      The function f is continuous for all real numbers and f (2) = (2 − 2) 4/5  = 0. Use
                                                      Form (1) of the definition of the derivative to find the two one-sided limits at 2.




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