Răspuns:
Explicație pas cu pas:
f : R --> R ; f(x) = ax+b
a) A(1 , 0) ; B(0 , 1) ∈ Gf <=>
f(1) = 0 ; f(0) = 1 => a+b = 0 ; a·0 + b = 1 => b = 1 => a = -1
f(x) = -x+1
b) A(-1 , -3) ; B(2 , 3) ∈Gf <=>
f(-1) = -3 ; f(2) = 3 =>
-a+b = -3 I·(-1) => a-b = 3 (1)
2a+b = 3 (2)
Din (2)+(1) => 3a = 6 => a = 2 => b = -1 => f(x) = 2x-1
c) A(√2 , 4) ; B(0 , 3)∈Gf <=>
f(√2) = 4 ; f(0) = 3
f(√2) = a√2+b = 4
f(0) = b = 3 => a = 1/√2 => a = √2/2 =>f(x) = √2/2 ·x +3
d) A(√2 , 2) ; B(√3 , √6) ∈Gf <=>
f(√2) = 2 ; f(√3) = √6 =>
a√2+b = 2
a√3+b = √6
a√3 - a√2 = √6-2 =>
a = (√6-2) / (√3-√2)
a = (√6-2)·(√3+√2)/(3-2)
a = √18+√12-√6-2√2
a = 3√2+2√3-√6-2√2
a = √2+2√3-√6
b = 2-a√2
b = 2-2-2√6+√12
b = -2√6+2√3 => f(x) = (√2+2√3-√6)·x +(2√3-2√6)
#copaceibrainly
