Question

Помогитееее!
Пусть f(x) = 2 - корень(1-x) ; g(x) = 1+2x / 3+x
Найдите область определения функции:
а)y=f(x)+g(x)
б)y=f(x) / g(x)

Answer 1

$f(x) = 2 - sqrt{1-x} \ g(x) = frac{1+2x}{3+x}$

$f(x) +g(x) = 2 - sqrt{1-x}+ frac{1+2x}{3+x} \ left { {{1-x geq 0} atop {3+x neq 0}} right. \ left { {{x leq 1} atop {x neq -3}} right. \ xin(-infty; -3)cup(-3;1]$

$f(x) = 2 - sqrt{1-x} \ g(x) = frac{1+2x}{3+x} \ frac{f(x)}{g(x)} = frac{(2 - sqrt{1-x})(3+x)}{1+2x} \ left { {{1-x geq 0} atop {1+2x neq 0}} right. \ left { {{x leq 1} atop {x neq -0.5}} right. \ xin(-infty; -0.5)cup(-0.5;1]$

Answer 2

a) $y=2- sqrt{1-x}+ frac{1+2x}{3+x}$
$1-x geq 0$
$x leq 1$
$3+x neq 0$
$x neq -3$
D(y)=(-∞;-3)U(-3;1]

2) $y= frac{2- sqrt{1-x}*(3+x)}{1+2x}$
$1-x geq 0$
$x leq 1$
$1+2x neq 0$
$x neq -frac{1}{2}$

D(y)= (-∞;-0,5)U(-0,5;1]