Whakaoti i te f^-1(x)=2x^2+1/sqrt{x}

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

Whakaoti mō f

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

$$x>0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakaraupapatia anō ngā kīanga tau.

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

Tē taea kia ōrite te tāupe f ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te f.

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

Whakamahia te āhuatanga tohatoha hei whakarea te fx^{-\frac{1}{2}} ki te 2x^{2}+1.

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

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te -\frac{1}{2} me te 2 kia riro ai te \frac{3}{2}.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Whakaraupapatia anō ngā kīanga tau.

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

Pahekotia ngā kīanga tau katoa e whai ana i te f.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

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

Mā te whakawehe ki te 2x^{\frac{3}{2}}+x^{-\frac{1}{2}} ka wetekia te whakareanga ki te 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

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

Whakawehe x ki te 2x^{\frac{3}{2}}+x^{-\frac{1}{2}}.

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

Tē taea kia ōrite te tāupe f ki 0.

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