Combinations of functions

COMBINATIONS OF FUNCTIONS

Two functions ƒ and g can be combined to form new functions f+g,f‒g. fg. and f/g in a manner similar to the way we add, subtract multiply, and divide real numbers. The sum and difference functions are defined by

(f + g)(x) = f(x) + g(x)

(fg)(x) = f(x) − g(x)

If the domain of ƒ is A and the domain of g is B, then the domain of ƒ + g is the intersection A∩B because both f(x) and g(x) have to be defined. For example, the domain of f(x) = √x is A = [0, ∞) and the domain of g(x) = √[2‒x] is B = (‒∞, 2]. So the domain of (f + g)(x) = √x + √[2‒x] is A∩B = [0,2].

Similarly, the quotient functions are defined by

 (fg)(x) = f(x)g(x)

(f/g) (x) = f(x) / g(x)

The domain of fg is A∩B, but we can't divide by 0 and so the domain of f/g is {x ϵ A∩B| g(x) ≠ 0). For instance, if f(x) = x2 and g(x) = x‒1, then the domain of the rational function (f/g)(x) = x2/(x‒1) is {x|x ≠ 1}, or (‒∞,1) U (1, ∞).

PROBLEMS UNDER COMBINATION OF FUNCTIONS

Example 14. Let f(x) = 1 +√[x‒2] and g(x) = x‒3. Find the domains and formulas for the functions i) f+g, ii) ƒg, iii) f‒g, iv) f/g, and v) 7f.

Solution: First, we will find the formulas and then the domains.

i) (f + g)(x) = f(x) + g(x)

= (1 + √[x‒2])+(x − 3)

= x ‒ 2 + √(x‒2

ii) (ƒ‒g)(x) = f(x) ‒ g(x)

 = (1 + √(x – 2)) ‒ (x − 3)

= 4‒x + √[x‒2])

iii) (fg)(x) = f(x)g(x)

 = (1 + √[x‒2])(x‒3)

iv) (f/g)(x) = f(x)/g (x) = { 1+√[x‒2] } / {x‒3}

v) (7ƒ)(x) = 7f(x) = 7 + 7√(x‒2)

The domains of ƒ and g are [2, +∞) and (‒∞, +∞), respectively (their natural domains). Thus, it follows from the definition, namely,

[2, + ∞) ∩ (‒∞, +∞) = [2, +∞)

Moreover, since g(x) = 0 if x = 3, the domain of f/g is [2, +∞) with x = 3 removed, namely,

[2,3) U (3, +∞)

Finally, the domain of 7f is the same as the domain of f.

Example 15. Show that if f(x) = √x. g(x) = √x, and h(x) = x, then the domain of f g is not the same as the natural domain of h.

Solution: The natural domain of h(x) = x is (‒∞, +∞).

Note that (fg)(x) = √x√x = x = h(x) on the domain of fg.

The domains of both ƒ and g are [0, +∞), so the domain of f g is

[0, +∞) ∩ [0, +∞0) = [0, +∞)

Since the domains of fg and h are different, it would be misleading to write (fg)(x) = x without including the restriction that this formula holds only for x ≥ 0.

Example 16. For the graphs of y = √x and y = 1/x, make a sketch that shows the general shape of the graph of y = √x+1/x for x≥ 0.

Solution:

To add the corresponding y‒values of y = √x and y = 1/x graphically, just imagine them to be "stacked" on top of one another.