Whakaoti i te f^-1(x)=2x^2-5x+1

$$f ^ { - 1 } ( x ) = 2 x ^ { 2 } - 5 x + 1$$

Whakaoti mō f

$$f=-\frac{x}{-2x^{2}+5x-1}$$

$$x\neq 0\text{ and }x\neq \frac{\sqrt{17}+5}{4}\text{ and }x\neq \frac{5-\sqrt{17}}{4}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakaraupapatia anō ngā kīanga tau.

$$\frac{1}{f}x=2x^{2}-5x+1$$

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.

$$1x=2x^{2}f-5xf+f$$

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

$$2x^{2}f-5xf+f=1x$$

Whakaraupapatia anō ngā kīanga tau.

$$2fx^{2}-5fx+f=x$$

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

$$\left(2x^{2}-5x+1\right)f=x$$

Whakawehea ngā taha e rua ki te 2x^{2}-5x+1.

$$\frac{\left(2x^{2}-5x+1\right)f}{2x^{2}-5x+1}=\frac{x}{2x^{2}-5x+1}$$

Mā te whakawehe ki te 2x^{2}-5x+1 ka wetekia te whakareanga ki te 2x^{2}-5x+1.

$$f=\frac{x}{2x^{2}-5x+1}$$

Tē taea kia ōrite te tāupe f ki 0.

$$f=\frac{x}{2x^{2}-5x+1}\text{, }f\neq 0$$

Whakaoti mō x (complex solution)

$$x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}$$

$$x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\neq 0$$

Whakaoti mō x

$$x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}$$

$$x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\leq \frac{-2\sqrt{2}-5}{17}\text{ or }\left(f\neq 0\text{ and }f\geq \frac{2\sqrt{2}-5}{17}\right)$$