Jawaban:

no 1)

f(x) = x² - 1

a) f(-3) , f(-1), f(0), f(2), f(3)

f(-3) = (-3)² - 1

= 9-1

= 8

f(-1) = (-1)² - 1

= 1 - 1

= 0

f(0) = (0)² - 1

= -1

f(2) = (2)² - 1

= 4 - 1

= 3

f(3) = (3)² - 1

= 9 - 1

= 8

b) jika D = {xI -3≤x≤3, x∈R}

D = {-3,-2,-1,0,1,2,3}

f(-3) = (-3)² - 1

= 9-1

= 8

f(-2) = (-2)² - 1

= 4 - 1

= 3

f(-1) = (-1)² - 1

= 1 - 1

= 0

f(0) = (0)² - 1

= -1

f(1) = 1² - 1

= 1 - 1

= 0

f(2) = (2)² - 1

= 4 - 1

= 3

f(3) = (3)² - 1

= 9 - 1

= 8

c) a jika f(a) = 3

f(a) = a² - 1

3 = a² - 1

a² = 4

a = 2 atau -2

2) range = {x|-2< x ≤ 4}

range = {-1,0,1,2,3,4}

a) g(x) = 4x - 2

g(-1) = 4(-1) -2

= -4-2

= -6

g(0) = 4(0) - 2

= 0-2

= -2

g(1) = 4(1) -2

= 4-2

= 2

g(2) = 4(2) - 2

= 8-2

= 6

g(3) = 4(3) -2

= 12-2

= 10

g(4) = 4(4) -2

= 16-2

= 14

range = {-4, -2, 2, 6, 10, 14}

b) g(x) = x²+4

g(-1) = (-1)² + 4

= 1+4

= 5

g(0) = (0)² + 4

= 4

g(1) = 1² + 4

= 5

g(2) = 2² + 4

= 4 + 4

= 8

g(3) = 3² + 4

= 9 + 4

= 13

g(4) = 4² + 4

= 16 + 4

= 20

range = {5, 4, 5, 8, 13, 20}

3)

a) h(x) = 2x³ + x

h(1) = 2(1)³ + 1

= 2+1

= 3

h(-1) = 2(-1)³ + (-1)

= -2 -1

= -3

-h(1) = -3

fungsi genap jika h(x) = h(-x) , dan fungsi ganjil jika h(x) = -h(x)

ternyata hasil cek menunjukkan tidak dua-duanya

b) h(x) = x² - 8x

h(1) = 1² - 8(1)

= 1-8

= -7

h(-1) = (-1)² -8(-1)

= 1 + 8

= 9

-h(x) = -(-7)

= 7

hasil cek menunjukkan tidak kedua-duanya

4) fungsi surjektif adalah fungsi jika daerah kodomain habis, maksudnya semua daerah kodomain dihubungi oleh daerah domain, jika soal dalam bentuk fungsi dan tidak ada daerah domain dan kodomain, tidak dapat menyebutkan apakah fungsi itu surjektif atau bijektif atau injektif

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