How do you find the domain and range of $f(x)= sqrt(x^2-x-6)$ ?

Question

23200 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-17T05:47:26

Domain is $(-oo,-2[ uu]3,+oo)$
Range is $[0,+oo)$

Since under square root the polynomial must be positive, the domain is obtained by solving:

$x^2-x-6>=0$

you can obtain the zeroes of the polynomial by solving the associated equation:

$x^2-x-6=0$

$x=(1+-sqrt(1-4*(-6)))/2$

$x=(1+sqrt(25))/2$

$x=(1+-5)/2$

$x=-2 and x=3$

so the disequation is solved in the external intervals:

$x<-2 and x>3$
the the domain is:

$(-oo,-2[ uu]3,+oo)$

Since f(x) is positive, due to the result of square root, the range includes all positive real numbers: $x>=0$

How do you find the domain and range of f(x)= sqrt(x^2-x-6)f(x)= sqrt(x^2-x-6)?