Home › Functions and graphs

How to find (f + g), (f − g), (f · g), (f / g) and their domains

The rule: combine the formulas, intersect the domains

For a sum, a difference or a product you simply do the algebra:

(f \pm g) (x) = f (x) \pm g (x)

(f \cdot g) (x) = f (x) \cdot g (x)

and the domain is the intersection of the two original domains - a point has to be allowed for both functions. The quotient adds one more condition:

(\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}, g (x) \neq = 0

Here comes the trap worth remembering: when the fraction simplifies, the cancelled factor does not come back to life. If $g$ vanishes at a point, that point stays out of the domain even though the simplified formula looks perfectly healthy there.

Problem

Perform the operation, then find the domain, for

f (x) = x^{2} - 16, g (x) = x^{2} + 6 x + 8

Find $(f + g) (x)$ , $(f - g) (x)$ , $(f \cdot g) (x)$ , $(\frac{f}{g}) (x)$ and $(\frac{g}{f}) (x)$ , and state the domain of each.

Answer

(f + g) (x) = 2 x^{2} + 6 x - 8

and

(f - g) (x) = - 6 x - 24

, both on all of

R

(f \cdot g) (x) = x^{4} + 6 x^{3} - 8 x^{2} - 96 x - 128

, also on all of

R

(\frac{f}{g}) (x) = \frac{x - 4}{x + 2}

with

x \neq = - 4

and

x \neq = - 2

(\frac{g}{f}) (x) = \frac{x + 2}{x - 4}

with

x \neq = - 4

and

x \neq = 4

Checked by hand: every expansion and every domain was recomputed.

Step-by-step solution

1. Domains of the two functions

Both $f$ and $g$ are polynomials, and a polynomial is defined for every real number:

D (f) = D (g) = (- \infty, \infty)

It also helps to factor them straight away - the factors are what the quotients will need:

f (x) = x^{2} - 16 = (x - 4) (x + 4)

g (x) = x^{2} + 6 x + 8 = (x + 4) (x + 2)

2. Sum and difference

(f + g) (x) = (x^{2} - 16) + (x^{2} + 6 x + 8)

(f + g) (x) = 2 x^{2} + 6 x - 8

(f - g) (x) = (x^{2} - 16) - (x^{2} + 6 x + 8)

(f - g) (x) = x^{2} - 16 - x^{2} - 6 x - 8 = - 6 x - 24

Both are polynomials again, and the intersection of the two domains is everything:

D : (- \infty, \infty)

3. Product

(f \cdot g) (x) = (x^{2} - 16) (x^{2} + 6 x + 8)

= x^{4} + 6 x^{3} + 8 x^{2} - 16 x^{2} - 96 x - 128

(f \cdot g) (x) = x^{4} + 6 x^{3} - 8 x^{2} - 96 x - 128

D : (- \infty, \infty)

4. The quotient f / g

(\frac{f}{g}) (x) = \frac{x ^{2} - 16}{x ^{2} + 6 x + 8} = \frac{( x - 4 ) ( x + 4 )}{( x + 4 ) ( x + 2 )}

= \frac{x - 4}{x + 2}

The denominator of the original fraction vanishes when $(x + 4) (x + 2) = 0$ , that is at $x = - 4$ and $x = - 2$ . Both have to leave the domain - including $x = - 4$ , whose factor cancelled:

D : (- \infty, - 4) \cup (- 4, - 2) \cup (- 2, \infty)

At $x = - 4$ the simplified formula would happily return $\frac{- 8}{- 2} = 4$ , but the original quotient is $\frac{0}{0}$ there - undefined. On a graph this shows up as a hole.

5. The quotient g / f

(\frac{g}{f}) (x) = \frac{x ^{2} + 6 x + 8}{x ^{2} - 16} = \frac{( x + 4 ) ( x + 2 )}{( x - 4 ) ( x + 4 )}

= \frac{x + 2}{x - 4}

Now the denominator is $f$ , which vanishes at $x = 4$ and $x = - 4$ :

D : (- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)

Same story as before: $x = - 4$ is excluded because of the original denominator, even though the cancelled form says nothing about it.

Add to Chrome - solve your own

Screenshot any problem on your screen and get the steps. Chrome on a computer; free.