Question

If f(x) = log_e (1 – x) and g(x) = [x], then determine each of the following functions:
(i) f + g
(ii) fg

(iii) f/g
(iv) g/f 

Also, find (f + g) (–1), (fg) (0), (f/g) (1/2) and (g/f) (1/2).

Answer 1

Given as

f(x) = log_e (1 – x) and g(x) = [x]

As we know, f(x) takes real values only when 1 – x > 0

1 > x

x < 1, ∴ x ∈ (–∞, 1)

The domain of f = (–∞, 1)

Similarly, g(x) is defined for all real numbers x.

The domain of g = [x], x ∈ R

= R

(i) f + g

As we know, (f + g) (x) = f(x) + g(x)

(f + g) (x) = log_e (1 – x) + [x]

The domain of f + g = Domain of f ∩ Domain of g

The domain of f + g = (–∞, 1) ∩ R

= (–∞, 1)

∴ f + g: (–∞, 1) → R is given by (f + g) (x) = log_e (1 – x) + [x]

(ii) fg

As we know, (fg) (x) = f(x) g(x)

(fg) (x) = log_e (1 – x) × [x]

= [x] log_e (1 – x)

The domain of fg = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

∴ fg: (–∞, 1) → R is given by (fg) (x) = [x] log_e (1 – x)

(iii) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = log_e (1 – x)/[x]
The domain of f/g = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

However, (f/g) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when [x] = 0.

Now, we have, [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

If  0 ≤ x < 1, (f/g) (x) will be undefined as the division result will be indeterminate.

The domain of f/g = (–∞, 1) – [0, 1)

= (–∞, 0)

∴ f/g: (–∞, 0) → R is given by (f/g) (x) = log_e (1 – x)/[x] 

(iv) g/f

As we know, (g/f) (x) = g(x)/f(x) 

(g/f) (x) = [x]/log_e (1 – x)

However, (g/f) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when log_e (1 – x) = 0.

log_e (1 – x) = 0 ⇒ 1 – x = 1 or x = 0

If x = 0, (g/f) (x) will be undefined as the division result will be indeterminate.

The domain of g/f = (–∞, 1) – {0}

= (–∞, 0) ∪ (0, 1)

∴ g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f) (x) = [x]/log_e (1 – x) 

(a) Here, we need to find (f + g) (–1).

We have, (f + g) (x) = log_e (1 – x) + [x], x ∈ (–∞, 1)

By substituting x = –1 in the above equation, we get

(f + g)(–1) = log_e (1 – (–1)) + [–1]

= log_e (1 + 1) + (–1)

= log_e2 – 1

∴ (f + g) (–1) = log_e2 – 1

(b) Here, we need to find (fg) (0).

We have, (fg) (x) = [x] log_e (1 – x), x ∈ (–∞, 1)

By substituting x = 0 in the above equation, we get

(fg) (0) = [0] log_e (1 – 0)

= 0 × log_e1

∴ (fg) (0) = 0

(c) Here, we need to find (f/g) (1/2) 

We have, (f/g) (x) = log_e (1 – x)/[x], x ∈ (–∞, 0)

However, 1/2 is not in the domain of f/g.

∴ (f/g) (1/2) does not exist.

(d) Here, we need to find (g/f) (1/2) 

We have, (g/f) (x) = [x]/log_e (1 – x), x ∈ (–∞, 0) ∪ (0, ∞)

By substituting x = 1/2 in the above equation, we get

(g/f) (1/2) = [x]/log_e (1 – x)

= (1/2)/log_e (1 – 1/2)

= 0.5/log_e (1/2)

= 0/log_e (1/2)

= 0

Thus, (g/f) (1/2) = 0