f(x)=5x^3+4x^2+8x+1text{when}x=2 eching

$$f ( x ) = 5 x ^ { 3 } + 4 x ^ { 2 } + 8 x + 1 \text { when } x = 2$$

f, x uchun yechish

$$x=2$$

$$f = \frac{73}{2} = 36\frac{1}{2} = 36,5$$

Yechim qadamlari

Birinchi tenglamani yeching. Tenglamaga o‘zgaruvchining ma’lum qiymatlarini joylang.

$$f\times 2=5\times 2^{3}+4\times 2^{2}+8\times 2+1$$

3 daraja ko‘rsatkichini 2 ga hisoblang va 8 ni qiymatni oling.

$$f\times 2=5\times 8+4\times 2^{2}+8\times 2+1$$

40 hosil qilish uchun 5 va 8 ni ko'paytirish.

$$f\times 2=40+4\times 2^{2}+8\times 2+1$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$f\times 2=40+4\times 4+8\times 2+1$$

16 hosil qilish uchun 4 va 4 ni ko'paytirish.

$$f\times 2=40+16+8\times 2+1$$

56 olish uchun 40 va 16'ni qo'shing.

$$f\times 2=56+8\times 2+1$$

16 hosil qilish uchun 8 va 2 ni ko'paytirish.

$$f\times 2=56+16+1$$

72 olish uchun 56 va 16'ni qo'shing.

$$f\times 2=72+1$$

73 olish uchun 72 va 1'ni qo'shing.

$$f\times 2=73$$

Ikki tarafini 2 ga bo‘ling.

$$f=\frac{73}{2}$$

Tizim hal qilindi.

$$f=\frac{73}{2} x=2$$