If the rational function f and g are defined by rational epressions f(x) and g(x), then the product f*g is defined by f(x) and g(x) in simplest form.

Question

If the rational functions f and g are defined by rational expressions f(x) and g(x), then the product f*g is defined by f(x) * g(x) in simplest form, and the quotient f/g is defined by f(x)/g(x) in simplest form. Find rational functions f and g that meet the given conditions:

a) the zeros of f are 1 and 2 : the zeros of g are -2 and -3, the zeros of f*g are 1 and 3
b) the function f has one zero, g has one zero and f*g has no zeros
c) the function f has no zeros, g has no zeros and f/g has two zeros

Answer 1

For a) did you mean the zeroes of f*g are 1 and -3? If function g has a zero of -3 then function f * g should also have a zero of -3 because f(-3) * g(-3) = f(-3) * 0 = 0. [well this isn't quite technically true but I still think it's what you meant?]

So I'm going to assume you need f*g to have zeroes of 1 and -3.

 

Ok, so let's build the functions up...the simplest rational function f that has zeroes 1 and 2 would be:

f(x) = (x - 1)(x - 2)

 

The simplest rational function g that has zeroes -2 and -3 would be:

g(x) = (x + 2)(x + 3)

 

If we look at what f(x) * g(x) is right now it would be:

f*g(x) = (x - 1)(x - 2)(x + 2)(x + 3)

 

We would like f(x) * g(x) to keep the zeroes 1 and -3 but somehow get rid of the zeroes of 2 and -2.

We can do this by adding denominators to the functions so that when we multiply f and g some of the factors are canceled out. We use a denominator in f to cancel the zero of x = -2 coming from g and we use a denominator in g to cancel the zero of x = 2 coming from f.

 

Our new functions would look like this:

f(x) = (x - 1)(x - 2) / (x + 2)           The zeroes of f are still 1 and 2.

g(x) = (x + 2)(x + 3) / (x - 2)        The zeroes of g are still -2 and -3

 

f(x)*g(x) = (x - 1)(x - 2)(x + 2)(x + 3) / [ (x + 2)(x - 2) ]

We can cancel out the (x + 2) and (x - 2) factors and be left with:

f(x)*g(x) = (x - 1) (x + 3) while still noting that f*g is undefined at -2 and 2 because it originally had those factors in the denominator.

 

b) simply let f = (x+2) / (x+3) and g = (x+3) / (x+2) or anything similar. f * g will therefore be undefined at x = -2 and -3 are have no zeroes there.

 

c) let f = 1 and g = 1 / [(x + 1) (x + 2)]

f/g = (x + 1)(x + 2)

The division bumps the two factors in the numerator are therefore f /g will have zeroes of -1 and -2.