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How to add, subtract, multiply and divide functions when one has a root

The rule: combine the formulas, then intersect the domains

Adding, subtracting or multiplying two functions is just algebra on their formulas:

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

The domain, though, is not inherited from the simplified answer - it is the intersection of the two original domains: a value must be legal for $f$ and for $g$ at the same time. For the quotient there is one more condition on top of that:

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

So a root in $f$ keeps its restriction in every single combination, even where the root is no longer visible in the final formula.

Problem

For the functions

f (x) = 3 x, g (x) = 5 x - 3

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

Answer

(f + g) (x) = 3 x + 5 x - 3

and

(f - g) (x) = 3 x - 5 x + 3

, both on

[0, \infty)

(f \cdot g) (x) = (5 x - 3) 3 x

[0, \infty)

(\frac{f}{g}) (x) = \frac{3 x}{5 x - 3}

[0, \frac{3}{5}) \cup (\frac{3}{5}, \infty)

Checked by hand: every combination and every domain recomputed.

Step-by-step solution

1. Domains of the two functions first

The root in $f$ needs a non-negative inside:

3 x \geq 0 \Rightarrow x \geq 0, D (f) = [0, \infty)

The linear $g$ has no restrictions at all:

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

Their intersection - the base domain for every combination below - is therefore $[0, \infty)$ .

2. Sum and difference

(f + g) (x) = 3 x + 5 x - 3

(f - g) (x) = 3 x - (5 x - 3) = 3 x - 5 x + 3

Mind the bracket in the difference: the minus sign flips both terms of $g$ . Domain in both cases:

D : [0, \infty)

3. Product

(f \cdot g) (x) = 3 x (5 x - 3) = (5 x - 3) 3 x

Nothing new is forbidden, so the domain is again the intersection:

D : [0, \infty)

4. Quotient - the one with the extra condition

(\frac{f}{g}) (x) = \frac{3 x}{5 x - 3}

On top of $x \geq 0$ the denominator must not vanish:

5 x - 3 = 0 \Rightarrow x = \frac{3}{5}

That point sits inside $[0, \infty)$ , so it has to be cut out:

D : [0, \frac{3}{5}) \cup (\frac{3}{5}, \infty)

How to check yourself

Test the boundary values. At $x = 0$ the sum gives $0 + 0 - 3 = - 3$ - defined, so $0$ belongs. At $x = - 1$ the root is $- 3$ - not real, so negatives are rightly out. At $x = \frac{3}{5}$ the quotient becomes $\frac{1.8}{0}$ - undefined, exactly the point we removed.

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