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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`f(x)=sqrt(x^2-x-6)`

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.

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

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

`x2=[+1+sqrt(1+24)]/2=3`

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

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

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

`[0,+oo)`

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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