f^-1(x)=2x^2+1/sqrt{x} eching

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

f uchun yechish

$$f=\frac{x^{\frac{3}{2}}}{2x^{2}+1}$$

$$x>0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Shartlarni qayta saralash.

$$\frac{1}{f}x=\frac{2x^{2}+1}{\sqrt{x}}$$

f qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini f ga ko'paytirish.

$$1x=fx^{-\frac{1}{2}}\left(2x^{2}+1\right)$$

fx^{-\frac{1}{2}} ga 2x^{2}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$1x=2fx^{-\frac{1}{2}}x^{2}+fx^{-\frac{1}{2}}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. -\frac{1}{2} va 2 ni qo‘shib, \frac{3}{2} ni oling.

$$1x=2fx^{\frac{3}{2}}+fx^{-\frac{1}{2}}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2fx^{\frac{3}{2}}+fx^{-\frac{1}{2}}=1x$$

Shartlarni qayta saralash.

$$2fx^{\frac{3}{2}}+x^{-\frac{1}{2}}f=x$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(2x^{\frac{3}{2}}+x^{-\frac{1}{2}}\right)f=x$$

Tenglama standart shaklda.

$$\left(2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}\right)f=x$$

Ikki tarafini 2x^{\frac{3}{2}}+x^{-\frac{1}{2}} ga bo‘ling.

$$\frac{\left(2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}\right)f}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}=\frac{x}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}$$

2x^{\frac{3}{2}}+x^{-\frac{1}{2}} ga bo'lish 2x^{\frac{3}{2}}+x^{-\frac{1}{2}} ga ko'paytirishni bekor qiladi.

$$f=\frac{x}{2x^{\frac{3}{2}}+\frac{1}{\sqrt{x}}}$$

x ni 2x^{\frac{3}{2}}+x^{-\frac{1}{2}} ga bo'lish.

$$f=\frac{x^{\frac{3}{2}}}{2x^{2}+1}$$

f qiymati 0 teng bo‘lmaydi.

$$f=\frac{x^{\frac{3}{2}}}{2x^{2}+1}\text{, }f\neq 0$$