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

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

Whakaoti mō f

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

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakaraupapatia anō ngā kīanga tau.

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

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.

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

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

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

Whakaraupapatia anō ngā kīanga tau.

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

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

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

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

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

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

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

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

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

Tē taea kia ōrite te tāupe f ki 0.

$$f=x\left(\sqrt{x^{2}+1}+x\right)\text{, }f\neq 0$$

Whakaoti mō x

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

$$f>-\frac{1}{2}\text{ and }f\neq 0$$