Originally Posted by
Riley123 What is the domain of these problems??
R(x) = 5x squared over 10-x
Q(x) = -x(8-x) over 3xsquared + 7x+2
H(x) = x-7 over 2x to the 4th +7
These would be clearer if (remaining with ASCII) you write them:
R(x) = 5 x^2/(10-x)
Q(x) = -x (8-x)/(3 x^2 + 7 x + 2)
H(x) = (x-7)/(2 x^4 + 7)
where I have to make several guesses at what you actual mean (note
here ^ denotes raising to a power, so x^2 is x squared).
Better yet would be:
$\displaystyle R(x)\ =\ \frac{5\ x^2}{(10-x)}$
$\displaystyle Q(x)\ =\ \frac{-x\ (8-x)}{(3x^2\ +\ 7x\ +\ 2)}$
$\displaystyle H(x)\ =\ \frac{(x-7)}{(2x^4\ +7)}$
The domain of a function is the set on which the function is defined.
From the look of this problem it looks as through we are actually being
asked for the set of real numbers on which the functions are defined.
Though what follows would be equally applicable if we were working
with the complex numbers.
In all of these function definitions both the numerators and
denominators can be evaluated for all real numbers x, but
the functions are undefined when the denominators are zero.
In the first case the denominator is (10-x) which is zero when
x is equal to 10, so the domain of R is the set of all real number
except 10.
The rest are similar, in each case the domain is the set of all
of the reals except the points at which the denominator are
zero (it at all).
RonL