Argument x2 jest o 3 większy...- Zadanie 1: MATeMAtyka 1. Zakres podstawowy

Rozwiązanie

Argument x₂ jest o 3 większy od argumentu x₁, więc x₂=x₁+3.

$a) f (x_{2}) - f (x_{1}) = 2 x_{2} - 2 x_{1} = 2 (x_{2} - x_{1}) = 2 (x_{1} + 3 - x_{1}) = 2 \cdot 3 = 6$

$b) f (x_{2}) - f (x_{1}) = \frac{1}{3} x_{2} + 5 - (\frac{1}{3} x_{1} + 5) = \frac{1}{3} x_{2} + 5 - \frac{1}{3} x_{1} - 5 = \frac{1}{3} x_{2} - \frac{1}{3} x_{1} = \frac{1}{3} (x_{2} - x_{1}) = \frac{1}{3} (x_{1} + 3 - x_{1}) = \frac{1}{3} \cdot 3 = 1$

$c) f (x_{2}) - f (x_{1}) = - \frac{2}{3} x_{2} + \frac{1}{3} - (- \frac{2}{3} x_{1} + \frac{1}{3}) = - \frac{2}{3} x_{2} + \frac{1}{3} + \frac{2}{3} x_{1} - \frac{1}{3} = - \frac{2}{3} x_{2} + \frac{2}{3} x_{1} = - \frac{2}{3} (x_{2} - x_{1}) = - \frac{2}{3} (x_{1} + 3 - x_{1}) = - \frac{2}{3} \cdot 3 = - 2$

$d) f (x_{2}) - f (x_{1}) = 7 - 7 = 0$

$e) f (x_{2}) - f (x_{1}) = 3 x_{2} - 3 - (3 x_{1} - 3) = 3 x_{2} - 3 - 3 x_{1} + 3 = 3 x_{2} - 3 x_{1} = 3 (x_{2} - x_{1}) = 3 (x_{1} + 3 - x_{1}) =$ 3*3=9

$f) f (x_{2}) - f (x_{1}) = 5 x_{2} + 1 - (5 x_{1} + 1) = 5 x_{2} + 1 - 5 x_{1} - 1 = 5 x_{2} - 5 x_{1} = 5 (x_{2} - x_{1}) = 5 (x_{1} + 3 - x_{1}) = 5 \cdot 3 = 35$