39. $$ f(g(x))=\frac{\sqrt{1-x^{2}}}{2} $$ va $$ g(x)=2 x+1 $$ funksiyalar berilgan. a) $$ f(x) $$ funksiyaning aniqlanish sohasini toping.

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Berilgan funksiyalar:

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

a) $$ f(x) $$ funksiyasining aniqlanish sohasini topish uchun avval $$ g(x) $$ funksiyasining aniqlanish sohasini aniqlaymiz.

$$ g(x) = 2x + 1 $$ funksiyasi har qanday haqiqiy son uchun aniqlanadi, ya'ni $$ x \in \mathbb{R} $$.

Endi $$ f(g(x)) $$ funksiyasini ko'rib chiqamiz. $$ f(g(x)) $$ ni aniqlash uchun $$ \sqrt{1 - x^2} $$ ifodasining aniqlanish sohasini aniqlashimiz kerak.

Bu ifoda aniqlanishi uchun quyidagi shart bajarilishi kerak:

$$
1 - x^2 \geq 0
$$

Bu tengsizlikni yechamiz:

$$
1 \geq x^2
$$

Bu tengsizlikni yechish uchun $$ x $$ ni quyidagi ko'rinishda yozamiz:

$$
-\sqrt{1} \leq x \leq \sqrt{1}
$$

Shu bilan birga, $$ x $$ ning qiymatlari $$ -1 \leq x \leq 1 $$ oraliqda bo'lishi kerak.

Endi $$ g(x) $$ ni $$ x $$ o'rniga qo'yamiz:

$$
g(x) = 2x + 1
$$

Bu ifodani $$ -1 \leq g(x) \leq 1 $$ shartiga qo'yamiz:

1. $$ -1 \leq 2x + 1 \leq 1 $$

Bu tengsizlikni yechamiz:

- Birinchi qism: $$ -1 \leq 2x + 1 $$

$$
-1 - 1 \leq 2x \implies -2 \leq 2x \implies -1 \leq x
$$

- Ikkinchi qism: $$ 2x + 1 \leq 1 $$

$$
2x \leq 1 - 1 \implies 2x \leq 0 \implies x \leq 0
$$

Shunday qilib, $$ x $$ ning aniqlanish sohasini quyidagicha yozishimiz mumkin:

$$
-1 \leq x \leq 0
$$

Natijada, $$ f(x) $$ funksiyasining aniqlanish sohasini topdik:

**Javob:** $$ f(x) $$ funksiyasining aniqlanish sohasi $$ [-1, 0] $$.
$$ f(x) $$ funksiyasining aniqlanish sohasi $$[-1, 0]$$.

39. \( f(g(x))=\frac{\sqrt{1-x^{2}}}{2} \) va \( g(x)=2 x+1 \) funksiyalar berilgan. a) \( f(x) \) funksiyaning aniqlanish sohasini toping.

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