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Show that f(x) = loga x, 0 < a < 1 is a decreasing function for all x > 0.
Given:- Function f(x) = loga x , 0 < a < 1
Theorem:- Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(x) > 0 and solve this inequation.
For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.
Here we have,
f(x) = loga x, 0 < a < 1
As given 0 < a < 1
⇒ log(a) < 0
and for x > 0
Therefore f’(x) is
⇒ f’(x) < 0
Hence, condition for f(x) to be decreasing
Thus f(x) is decreasing for all x > 0