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Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan
                3.  Selesaikan soalan graf fungsi f(x) yang berikut.   SP 2.1.1     TP5
                  Solve the following questions of the function graph f(x).

                                        y                            (a)                     y
                                       4
                                               y = f(x)                                         y = f(x)
                                       3                                                    3
                                       2

                                    –2   0      5  6  x                                  –2  0     3  4  x
                                         –1                                                   –1

                       Rajah menunjukkan sebahagian graf fungsi f(x)      Rajah menunjukkan sebahagian graf fungsi f(x)
                       untuk –2 < x < 6. Berdasarkan kepada graf,       untuk –2 < x < 4. Berdasarkan kepada graf,
                              dx Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
                       The diagram shows a part of the function f(x) for –2 < x < 6.      The diagram shows a part of the function f(x) for  –2 < x < 4.
                       Based on the graph,                              Based on the graph,
                       (i)  cari/find had/lim f(x)                      (i)  cari / find f(–2)
                                    x → –2
                       (ii)  tentukan sama ada had  f(x) wujud atau     (ii)  tentukan  sama  ada  had  f(x)  wujud  atau
                                                                                                 x → 0
                           tidak. Jelaskan.    x → 0                         tidak. Jelaskan.
                           determine whether lim f(x) exists or not. Explain.        determine whether lim f(x) exists or not. Explain.
                                          x → 0                                            x→0
                         (i)  had/lim f(x) = 4                          (i)  f(–2) tidak tertakrif, maka tidak ada nilai.
                             x → -2                                          f(–2) is not defined, hence there is no value.

                       (ii)  Kiri/Left:  had/lim f(x) = –1              (ii)  Kiri/Left: had/lim f(x) = 3
                                     x → 0
                                                                                      x → 0
                           Kanan/Right:  had/lim f(x) = 2                    Kanan/Right: had/lim f(x) = 3
                                         x → 0
                                                                                         x → 0
                                  Nilai tidak sama, nilai  had f(x) tidak wujud.          Nilai sama, nilai had f(x) wujud.
                                                x → 0                                      x → 0
                           The values are not the same, lim  f(x) does not exist.
                                                 x → 0                       Same value, lim f(x) exists.
                                                                                     x → 0


                4.  Cari  dy  dengan menggunakan prinsip pertama bagi setiap fungsi y = f(x) yang berikut.   SP 2.1.2     TP3
                       dx
                      dy
                   Find   dx   by using the first principles for each of the following functions y = f(x).



                                                                            –1
                                                                    (a) y =
                            y  = 2x – 1                                     x
                        y + δy  = 2(x + δx) – 1     Tip Penting                         y + δy  =   –1
                    2x – 1 + δy  = 2x + 2δx – 1     Gunakan/ Use                                 (x + δx)
                                                                               (y + δy)(x + δx)  = –1
                   	       δy  = 2δx            dy   =  had/lim δy       yx + yδx + xδy + δx.δy  = –1

                                                dx
                                                    δx → 0 δx
                           δy   = 2                                                xδy + δx.δy  = –yδx
                           δx
                   Maka/Hence  dy   =  had/lim δy                                   (x + δx)δy  = –yδx

                                   δx → 0  δx                                              δy   =   1
                                 =  had/lim  2                                             δx    x(x + δx)
                                   δx → 0                                          dy   had/lim δy

                                 = 2                                    Maka/Hence  dx  =  δx → 0 δx
                                                                                      =  had/lim    1
                                                                                        δx → 0  x(x + δx)
                                                                                      =   1
                                                                                        x(x)

                                                                                      =  1
                                                                                        x 2






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         02 Hybrid PBD Mate Tamb Tg5.indd   20                                                                    09/11/2021   9:24 AM
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