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a) f(x)=3x+4
Df = R
b) f(x)=7/x-2
Df = R - {0} , bo w mianowniku nie może być zero
c) f(x)=5-x/x+3
Df = R - {0} , bo w mianowniku nie może być zero
d) f(x)=√x+3
Df = R+ , bo pod pierwistkiem kwadratowym nie moę być liczba ujemna
2. wyznacz jesli istnieja wszystkie miejsca zerowe:
a) f(x)=x+3
x + 3 = 0
x = -3
b) f(x)=-3x(x-2)(x+1)(5x-4)
-3x(x-2)(x+1)(5x-4) = 0
-3x = 0 lub x -2 = 0 lub x + 1 = 0 lub 5x -4 = 0
x = 0, lub x = 2, lub x = -1, lub x = 4/5
c) f(x)=3/x
brak miejsc zerowych
3. oblicz f(1), f(-¼), f(√2), f(√3-1)
a) f(x)=2x-3
f(1) = 2*1 -3 = 2 -3 = -1
f(-1/4) = 2*(-1/4) -3 = -1/2 - 3 = -3½
f(√2) = 2*√2 -3 = 2√2 -3
f(√3-1)= 2*(√3-1) -3 = 2√3 - 2 -3 = 2√3 -5
b) f(x)=3x²-4x-2
f(1)= 3*1² -4*1 -2 = 3 - 4 -2 = -3
f(-¼)= 3*(-1/4)² -4*(-1/4) -2 = 3/16 + 1 -2 = 19/16 - 32/16 = -13/16
f(√2)=3(√2)² -4*√2 -2 = 3*2 -4*√2 -2 = 4 - 4√2= 4(1 -√2)
f(√3-1)=3*(√3-1)² -4*(√3 -1) -2 = 3(3 -2√3 +1) -4√3 +4 -2 =3(4 -2√3) -4√3 +2=
= 12 -6√3 -4√3 +2 = 14 -10√3
c) f(x)= 4+x/(2-x)
f(1)= 4 + 1/(2 - 1)= 4 + 1 = 5
f(-¼)= 4 + (-1/4)/[2 -(-1/4)] = 4 + (-1/4)/[(9/4)]= 4 + (-1/4)*(4/9) = 4 -1/9= 3i 8/9
f(√2)=4 + √2/(2 -√2) = 4 + [√2 : (2 -√2]*[ (2 + √2)*( 2 + √2)] =
= 4 + √2(2 + √2):(4 -2) = 4 + √2(2 +√2) : 2 = 4 + √2 + 1 = 5 +√2
f(√3-1) =4 + (√3-1):[2 -(√3-1) ] = 4 +( √3-1):[ 2-√3 +1] = 4 +( √3-1):(3 - √3)=
= 4 + [( √3-1):(3 - √3)]*[(3 + √3):(3 + √3)]=
= 4 + [( √3-1)*(3 + √3 ] : ( 9 -3) =
= 4 + [ 3√3+ 3 -3 -√3] : 6 =
= 4 + [ 2√3]:6 = 4 + 1/3*√3