Find the Domain and Range of` f(x)= sqrt(x^2-x-6)` Write in set-builder notation and interval notation. x)=x2−x−6−−−−−−−−

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durbanville | High School Teacher | (Level 2) Educator Emeritus

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To find the domain and range of a quadratic function, consider

`f(x) = ax^2+bx+c` (standard form) or `f(x)=a(x-h)^2+k` (vertex form)

Note in this equation, the square root imposes some restrictions. We have a positive square root only (not `+-sqrt(x)` for example.) Note also that `y >= k`

From `f(x)=x^2-x-6`  we get `f(x)=(x+2)(x-3)`    

Establish which way the inequality will go:

`(x+2)(x-3).=0` (as it is positive)For this inequality to be true: `x<=-2`  and `x>=3` .Plug in values smaller than, in between and greater than these values to confirm this:

eg. when x = -4 = (-4+2)(-4-3) which = (-2)(-7). The answer is not important but the fact that it renders a positive result, means it is true.  

Therefore D={x: `x in RR| x<=-2 and x>=3` }

Therefore Range = [y: `y>=0`

Please take care not to ask multiple questions - either domain and range or say domain in set builder and interval. Please post any remaining questions separately. 

Ans: {x: `x in RR|x<=-2 and x>=3}`

and {y: `y in RR|y>=0}`


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valentin68 | College Teacher | (Level 3) Associate Educator

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To find the domain of the function one need to find the intervals on the real axis where the expression under the radical is zero or positive.


the roots of the equation `x^2-x-6=0` are `x1=[+1-sqrt(1+24)]/2=-2`


therefore the expression under the radical is positive or zero for `x<=-2 or x>=3`

Hence the domain of the function is

`{x: x<=-2 or x>=3}` or in interval notation


To find the range of the function one need to evaluate the limit of `f(x)`  as x aproaches the points `-oo,+oo,-2 and 3`

`lim_(x->oo)f(x)= lim_(x->-oo)f(x)=oo `

`lim_(x->-2)f(x)=lim_(x ->3)f(x) =0`

Therefore the range of the function is

`{x: x>=0}`  or in interval notation