Prove that the function f(x) = loga x is strictly increasing on] 0, �

1.	Prove that the function f(x) = loga x is strictly increasing on] 0, ∞ [when a > 1 and strictly decreasing on] 0, ∞ [ when 0 < a < 1.
Answer» It is given that f(x) = log_a x (a) By differentiating w.r.t x f’(x) = 1/x log a > 0 for x ∈ (0, ∞) and a > 1 Here f’(x) > 0 for x ∈ (0, ∞) and a > 1 Therefore, f’(x) is strictly increasing function. (b) We know that 1/x log a < 0 for x ∈ (0, ∞) where 0 < a < 1 So f’(x) < 0 Therefore, f’(x) is strictly decreasing on (0, ∞) and 0 < a < 1.

Prove that the function f(x) = loga x is strictly increasing on] 0, ∞ [when a > 1 and strictly decreasing on] 0, ∞ [ when 0 < a < 1.

Answer»

It is given that

f(x) = log_a x

(a) By differentiating w.r.t x

f’(x) = 1/x log a > 0 for x ∈ (0, ∞) and a > 1

Here f’(x) > 0 for x ∈ (0, ∞) and a > 1

Therefore, f’(x) is strictly increasing function.

(b) We know that

1/x log a < 0 for x ∈ (0, ∞) where 0 < a < 1

So f’(x) < 0

Therefore, f’(x) is strictly decreasing on (0, ∞) and 0 < a < 1.