Whakaoti i te f(x)=x^2+1quadg(x)=-4x-1

$$f ( x ) = x ^ { 2 } + 1 \quad g ( x ) = - 4 x - 1$$

Whakaoti mō g

$$g=-x-4-\frac{1}{x}$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te x^{2} mai i ngā taha e rua.

$$1gx=-4x-1-x^{2}$$

Whakaraupapatia anō ngā kīanga tau.

$$gx=-x^{2}-4x-1$$

He hanga arowhānui tō te whārite.

$$xg=-x^{2}-4x-1$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xg}{x}=\frac{-x^{2}-4x-1}{x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$g=\frac{-x^{2}-4x-1}{x}$$

Whakawehe -x^{2}-4x-1 ki te x.

$$g=-x-4-\frac{1}{x}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

$$x=\frac{-\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

Whakaoti mō x

$$x=\frac{\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}$$

$$x=\frac{-\sqrt{\left(g+2\right)\left(g+6\right)}-g-4}{2}\text{, }g\leq -6\text{ or }g\geq -2$$