Question

Sa se determine functia f:R->R, definita prin f(x)=ax+b

Answer 1

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)

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