Question

Find the domain and range of $y=\sqrt{x^2+6x-27}$

Answer 1

Given,

$\longrightarrow y=\sqrt{x^2+6x-27}$

We know the function $f(x)=\sqrt x$ is defined for $x\geq0.$

Thus if $y$ has to be defined,

$\longrightarrow x^2+6x-27\geq0$

$\longrightarrow x^2+9x-3x-27\geq0$

$\longrightarrow x(x+9)-3(x+9)\geq0$

$\longrightarrow (x+9)(x-3)\geq0$

$\Longrightarrow\underline{\underline{x\in(-\infty,\ -9]\cup[3,\ \infty)}}$

This is the domain.

Now,

$\longrightarrow y=\sqrt{x^2+6x-27}$

$\longrightarrow y^2=x^2+6x-27$

$\longrightarrow x^2+6x-27-y^2=0$

$\longrightarrow x^2+6x-(y^2+27)=0$

Now we've a quadratic equation in $x$ whose discriminant should be non - negative in order to get real values for $x.$

$\longrightarrow 6^2+4(y^2+27)\geq0$

$\longrightarrow 36+4y^2+108\geq0$

$\longrightarrow 4y^2+144\geq0$

$\longrightarrow y^2+36\geq0$

$\longrightarrow y^2\geq-36$

If the square of a real number is shown to be greater than a negative number, that means the square is always non - negative, since due to the fact that there does not exist any positive real number between zero and a negative number.

Therefore,

$\longrightarrow y^2\geq0$

$\Longrightarrow y\in\mathbb{R}\quad\quad\dots(1)$

We know the range of the function $f(x)=\sqrt x$ is that $f(x)\in[0,\ \infty).$

So, since $y=\sqrt{x^2+6x-27},$

$\longrightarrow y\in[0,\ \infty)\quad\quad\dots(2)$

Taking $(1)\land(2),$

$\longrightarrow y\in\mathbb{R}\cap[0,\ \infty)$

$\longrightarrow\underline{\underline{y\in[0,\ \infty)}}$

This is the range.