Question 28 of 82beginner🔧 ApplyShort Answer3 marks

Prove that the logarithmic function is increasing on (0, ∞).

Correct Answer

Let f(x)=log⁡xf(x) = \log x be defined on the interval (0,∞)(0, \infty).

Differentiating the function with respect to xx, we get the derivative as f′(x)=1xf'(x) = \frac{1}{x}.

For the interval (0,∞)(0, \infty), the value of xx is strictly positive (i.e., x>0x > 0). Consequently, the derivative f′(x)=1xf'(x) = \frac{1}{x} is also strictly positive for all xx in this interval.

According to the theorem derived from the Mean Value Theorem provided in the context, if the derivative of a function is positive in an interval, the function is increasing in that interval. Since f′(x)>0f'(x) > 0 for all x∈(0,∞)x \in (0, \infty), the logarithmic function is increasing on (0,∞)(0, \infty).

Exercise: EXERCISE 6.2 | Q: 10 | (Page 13)
For More Understanding

Explanation

The solution applies the theorem mentioned in the 'Proof (a)' and 'Remarks' sections of the provided context, which states that if f′(x)>0f'(x) > 0 for all xx in an interval, the function is increasing. Although the specific derivative of log⁡x\log x is not explicitly stated in the text chunks, it is the standard calculation required to solve Question 10 listed in the context. The proof relies on establishing that the derivative is positive for all x>0x > 0.

Solution Steps

  1. Step 1: Define the function f(x)=log⁡xf(x) = \log x on (0,∞)(0, \infty).

  2. Step 2: Calculate the derivative: f′(x)=1xf'(x) = \frac{1}{x}.

  3. Step 3: Analyze the sign of the derivative. Since x>0x > 0 in the given interval, f′(x)=1x>0f'(x) = \frac{1}{x} > 0.

  4. Step 4: Conclude using the theorem that since f′(x)>0f'(x) > 0 on the interval, the function is increasing.

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