Suppose f and g be two real functions continuous at a real number c. Then
(1) f + g is continuous at x = c.
(2) f – g is continuous at x = c.
(3) f . g is continuous at x = c.
(4) `(f/g)` is continuous at x = c ,
(provided g(c) ≠ 0).
Proof: We are investigating continuity of (f + g) at x = c.
Clearly it is defined at x = c. We have
`lim_(x->c)`(f + g)(x) = `lim_(x->c)`[f(x) + g(x)]
(by defination of f + g)
`lim_(x->c)`f(x) + `lim_(x->c)`g(x)
(by the theorem on limits)
= f(c) + g(c) (as f and g are continuous)
=(f+g)(c) (by defination of f + g)
Hence, f + g is continuous at x = c. Proofs for the remaining parts are similar and left as an exercise to the reader.
(i) As a special case of (3) above, if f is a constant function, i.e., f(x) = λ for some real number λ, then the function (λ . g) defined by (λ . g) (x) = λ . g(x) is also continuous. In particular if λ = – 1, the continuity of f implies continuity of – f.
(ii) As a special case of (4) above, if f is the constant function f(x) = λ, then the function `lambda/g` defined by `lambda/g(x)` = `lambda/g(x)` is also continuous wherever g(x) ≠ 0.
In particular, the continuity of g implies continuity of `1/g`
The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not.
Shaalaa.com | Continuity and Differentiability part 12 (Continuity composite function)
A function f (x) is defined as
f (x) = x + a, x < 0
= x, 0 ≤x ≤ 1
= b- x, x ≥1
is continuous in its domain.
Find a + b.
If f (x) is continuous on [–4, 2] defined as
f (x) = 6b – 3ax, for -4 ≤ x < –2
= 4x + 1, for –2 ≤ x ≤ 2
Show that a + b =`-7/6`