Question

Find the domain and range of the function $f\left(x\right)=\sqrt{x^2-25}$ and $f\left(x\right)=\sqrt{25-x^2}$

Answer 1

$f\left(x\right)=\sqrt{x^2-25}.$

$⟹x^2-25\ge 0$

$⟹(x-5)(x+5)\ge 0x\in (-\infty ,-5]\cup [5,\infty ).$

$⟹y=\sqrt{x^2-25}$

$⟹y^2+25=x^2$

$\therefore x=\pm \sqrt{y^2+25}$

$f\left(x\right)=\sqrt{x^2-25}.$

$⟹25-x^2\ge 0$

$⟹x^2-25\le 0$

$⟹(x+5)(x-5)\ge 0$

$x[-5,5].$

$f\left(x\right)=\sqrt{25-x^2}$

$⟹25-x^2\ge 0$

$⟹x^2-25\le 0$

$⟹x+5(x-5)\le 0$

$x\in (-5,5]$

$⟹y=\sqrt{25-x^2}$

$⟹y^2-25=-x^2$

$⟹x^2=25-y^2$

$x=\sqrt{25-y^2}$

Hence $25-y^2\ge 0(5-y)(5+y)\ge 0(y-5)(y+5)\le 0$

$\therefore y\in [-5,5]$

$y$ is defined for all values.

$\therefore y\in R.$