Prove that the logarithmic function is strictly increasing on \((0, ∞)\).
Prove that the logarithmic function is strictly increasing on \((0, ∞)\).
Solution and Explanation
The given function is f(x) = log x.
∴f'(x) = \(\frac 1x\)
It is clear that for x>0, f'(x) = \(\frac 1x\) >0.
Hence, f(x) = log x is strictly increasing in interval (0, ∞).
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Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
