Question

1. Oblicz wartości funkcji ƒ dla argumentów -1,0,½,√2,2 .
a) ƒ(x)= x kreska ułamkowa 2x+1
b) ƒ(x)= Ι2x-2Ι
c) ƒ(x)={ 0 dla x∈ (-∞,1>
x²-1 dla x∈ (1,2>
1 dla x∈ (2, ∞)
d)ƒ(x)={ x +6 dla x∈ (-∞,0>
x²+6 dla x∈ (0,³/₂>
x³+6 dla x∈ (³/₂,∞>



2. Punkty A,B,C,D należą do wykresu funkcji ƒ. Wyznacz a,b,c i d.
a)ƒ(x)=1 kreska ułamkowa 2x , A(1,a), B(2,b),C(c,2),D(d√2)
b)ƒ(x)= x²-1 kreska ułamkowa x-1, A(99,a), B(b,99),C(0,c),D(d,0)
c)ƒ(x)=√x²-6x+9, A(0,a),B(b,0), C(33,c),D(-33,d)
PILNE!!!!

Answer 1

Zad. 1
a) f(x) = x/(2x +1)
f(-1) = (-1)/[2*(-1) +1] = (-1)/[-2+1] = (-1)/(-1) = 1
f(0) = 0/(2*0+1) = 0/1 = 0
f(½) = (½)/(2*½+1) = (½)/(1+1) = (½)/2 = ¼
f(√2) = (√2)/(2√2+1) = [(√2)*(2√2 -1)]/[(2√2+1)(2√2-1)]=
(4 -√2)/(8-1) = (4 -√2)/7
f(2) = 2/(2*2+1) = 2/5

b) ƒ(x)= Ι2x-2Ι
f(-1) = I2*(-1)-2I = I-2-2I = I-4I = 4
f(0) = I2*0-2I = I0-2I = I-2I = 2
f(½) = I2*½-2I = I1-2I = I-1I = 1
f(√2) = I2*√2-2I = I2√2-2I = I2(√2-1)I = 2(√2-1)
f(2) = I2*2-2I = I4-2I = I2I = 2

c) ƒ(x)={ 0 dla x∈ (-∞,1>
x²-1 dla x∈ (1,2>
1 dla x∈ (2, ∞)
f(-1) = 0, bo x∈ (-∞,1>
f(0) = 0, bo x∈ (-∞,1>
f(½) = 0, bo x∈ (-∞,1>
f(√2) = x²-1 = (√2)²-1 = 2-1 = 1, bo x∈ (1,2>
f(2) = x²-1 = 2²-1 = 4-1 = 3, bo x∈ (1,2>

d)ƒ(x)={ x +6 dla x∈ (-∞,0>
x²+6 dla x∈ (0,³/₂>
x³+6 dla x∈ (³/₂,∞>
f(-1) = x+6 = -1+6 = 5, bo x∈ (-∞,0>
f(0) = x+6 = 0+6 = 6, bo x∈ (-∞,0>
f(½) = x²+6 = (½)²+6 = ¼+6 = 6¼, bo x∈ (0,³/₂>
f(√2) = x²+6 = (√2)²+6 = 2+6 = 8, bo x∈ (0,³/₂>
f(2) = x³+6 = 2³+6 = 8+6 = 14, bo x∈ (³/₂,∞>

Zad.2
a) ƒ(x)=1/(2x)
I) A(1,a)
f(1) = 1/(2*1) = ½
a = ½
II) B(2,b)
f(2) = 1/(2*2) = ¼
b = ¼
III) C(c,2)
f(c) = 2, 1/(2c) = 2, 2c = ½, c = ¼
IV) D(d,√2)
f(d) = √2, 1/(2d) = √2, 2d = 1/√2, d = 1/(2√2) = √2/4

b)ƒ(x)= (x²-1)/(x-1), x nie może być równe 1
x²-1 = (x-1)(x+1)
(x²-1)/(x-1) = [(x-1)(x+1)]/(x-1) = x +1
I)A(99,a)
f(99) = (x²-1)/(x-1) = x+1 = 99+1 = 100
a=100
II)B(b,99)
f(b) = 99, (b²-1)/(b-1) = b+1= 99, stąd b = 99 -1 = 98
b=98
III)C(0,c)
f(0) = (x²-1)/(x-1) = x+1 = 0+1 = 1
c=1
IV)D(d,0)
f(d) = 0, (d²-1)/(d-1) = d+1= 0, stąd d = 0-1 = -1
d=-1

c) ƒ(x)=√(x²-6x+9)
I)A(0,a)
f(0) = √(x²-6x+9) =√(0²-6*0+9) = √9 = 3
a = 3
II)B(b,0)
f(b) = √(b²-6b+9)= 0, czyli b²-6b+9 = 0, (b-3)²= 0, stąd
b = 3
III)C(33,c)
f(33) = √(x²-6x+9) =√(33²-6*33+9) = √900 = 30
c = 30
IV)D(-33,d)
f(-33) = √(x²-6x+9) =√((-33)²-6*(-33)+9) = √1296 = 36
d = 36