How would I find the domain and range of: k(x)= (1/(4-x^2)+sqrt(x-1)?

Find the domain and range for
`k(x)=1/(4-x^2)+sqrt(x-1)`
.

(1)The domain is the set of all
possible inputs. Generally you assume that the domain is all real
numbers (unless the problem involves discrete objects like the number of people) and
then find any restrictions. The restrictions include, but are not limited to, not
dividing by zero and not taking even roots of negative
numbers.

So in this case we check for division by zero; if
x=2 then the fraction has a zero in the denominator, so x=2 is
not in the domain.

We also
check for taking even roots of a negative number: if class="AM">`x<1` then we are taking a square root of a negative
number which we cannot do in the real numbers.So x<1 is not
in the domain.

Thus the domain is class="AM">`x>=1,x!= 2`,or class="AM">`[1,2)uu(2,oo)` .

(2) The range
is the set of all possible outputs. It helps to look at a graph. type="image/svg+xml"
src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,1/(4-x^2)+sqrt(x-1),null,3,1,0,5,black,1,none"/>

Notice
that as x approaches 2 from the left, the function grows without bound. This is because
for 1.75<x<2, 4-x^2 is smaller than one and approaching zero, so the
reciprocal gets arbitrarily large (goes to positive
infinity).

As x approaches 2 from the right, for
2<x<2.23 4-x^2 is negative and smaller than 1 in absolute value, so the
fraction gets arbitrarily small (goes to negative
infinity).

As x gets large, the fraction gets close to
zero, but the square root grows without bound.

So the range
of the function is all real numbers; the function takes on every value for y at some
point.

Thus the domain is class="AM">`1<=x<2 uu x>2` and the range is all
real numbers.