Determine h(x) = f(x) / g(x) and then state the domain and range of h(x). a)`f(x)= 2x-3` `g(x) = 3x+1` b) `f(x) = sqrt(x-4)` `g(x) = x+2` ` <br> `
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Determine `h(x)=(f(x))/(g(x)) ` , then determine the domain and range of h(x):
(a) f(x)=2x-3; g(x)=3x+1
`h(x)=(2x-3)/(3x+1) `
h(x) is a rational function (a function that is a ratio of polynomials with the polynomial in the denominator not identically zero,i.e. not zero for all values of x.)
The domain of a function is the set of all possible inputs. For a rational function (with no common factors in the numerator and denominator) the domain consists of all real numbers except any value(s) of x where the denominator is zero.
Here `3x+1=0 ==> x=-1/3 ` so the domain is all real numbers except `x=-1/3 ` . This can be written in a number of ways: ` ``D: x != -1/3 ` ,`D={x|x != -1/3} ` , `D=RR ` `-{-1/3} ` , etc...
The range of a rational function where the numerator and denominator have the same degree is the ratio of the leading coefficients (assuming the polynomials in the numerator and denominator are written in standard form.)
Here the range is `y != 2/3 ` .
Thus `h(x)=(2x-3)/(3x+1),"Domain":x !=-1/3,"Range":y != 2/3 `
The graph of h(x):
(b) `f(x)=sqrt(x-4);g(x)=x+2 `
`h(x)=sqrt(x-4)/(x+2) `
Here the domain of the numerator is `x>=4 ` (Assuming the function is defined over the real numbers.). The domain of h(x) is all numbers in the domain of f(x),g(x) such that `g(x) !=0 ` . Since the only number causing a denominator of zero is x=-2, and this is not in the domain of f(x), we have the domain of h(x) is all real numbers greater than or equal to 4.
The numerator is always greater than or equal to zero, while for x>4 the denominator is positive, so the range (or all possible outputs) is `y>=0 ` . But this function tends to zero as x increases without bound.
Using calculus we find the first derivative is zero at x=10; using the first derivative test we determine that this is a maximum.
The function is `h(x)=sqrt(x-4)/(x+2) ` ; the domain is `x>=4 ` while the range is `0<= y <= sqrt(6)/12 ` or approximately `0<=y<=0.20412415 `
The graph:
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