Analyzing Monotonicity of Functions using Derivatives

To determine if a function is increasing or decreasing over a specific interval, we examine the sign of its first derivative in that interval.

• If \(f'(x) > 0\) for all \(x\) in an interval, then \(f(x)\) is increasing in that interval.
• If \(f'(x) < 0\) for all \(x\) in an interval, then \(f(x)\) is decreasing in that interval.
• If \(f'(x) = 0\) for all \(x\) in an interval, then \(f(x)\) is constant in that interval.

Statement 1: \(f(x) = \ln x\) is increasing in \((0, \infty)\)

Let's find the first derivative of \(f(x) = \ln x\).

The derivative of \(\ln x\) is given by:

\(f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}\)

Now, we need to analyze the sign of \(f'(x) = \frac{1}{x}\) in the interval \((0, \infty)\). This interval includes all positive real numbers.

For any \(x \in (0, \infty)\), we know that \(x > 0\). Therefore, the reciprocal \(\frac{1}{x}\) must also be positive.

So, \(f'(x) = \frac{1}{x} > 0\) for all \(x \in (0, \infty)\).

Since the first derivative \(f'(x)\) is positive throughout the interval \((0, \infty)\), the function \(f(x) = \ln x\) is increasing in this interval.

Thus, Statement 1 is correct.

Statement 2: \(g(x)=e^x+e^{\frac{1}{x}}\) is decreasing in \((0, \infty)\)

Let's find the first derivative of \(g(x) = e^x + e^{\frac{1}{x}}\).

\(g'(x) = \frac{d}{dx}\left(e^x + e^{\frac{1}{x}}\right)\)

Using the sum rule and chain rule:

\(g'(x) = \frac{d}{dx}(e^x) + \frac{d}{dx}\left(e^{\frac{1}{x}}\right)\)

\(g'(x) = e^x + e^{\frac{1}{x}} \cdot \frac{d}{dx}\left(\frac{1}{x}\right)\)

The derivative of \(\frac{1}{x}\) is \(\frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2}\).

So, \(g'(x) = e^x + e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)\)

\(g'(x) = e^x - \frac{e^{\frac{1}{x}}}{x^2}\)

For \(g(x)\) to be decreasing in \((0, \infty)\), \(g'(x)\) must be less than 0 for all \(x \in (0, \infty)\). That is, we need \(e^x - \frac{e^{\frac{1}{x}}}{x^2} < 0\), or \(e^x < \frac{e^{\frac{1}{x}}}{x^2}\), or \(x^2 e^x < e^{\frac{1}{x}}\) for all \(x \in (0, \infty)\).

Let's examine the behavior of \(g'(x)\) in the interval \((0, \infty)\).

• As \(x \to 0^+\): \(e^x \to e^0 = 1\). \(\frac{1}{x} \to \infty\), so \(e^{\frac{1}{x}} \to \infty\). \(x^2 \to 0\). The term \(\frac{e^{\frac{1}{x}}}{x^2}\) is a very large positive number divided by a very small positive number, so it tends to \(\infty\). \(g'(x) = e^x - \frac{e^{\frac{1}{x}}}{x^2} \approx 1 - \infty \to -\infty\). So, for small positive \(x\), \(g'(x) < 0\).
• As \(x \to \infty\): \(e^x \to \infty\). \(\frac{1}{x} \to 0\), so \(e^{\frac{1}{x}} \to e^0 = 1\). \(x^2 \to \infty\). \(g'(x) = e^x - \frac{e^{\frac{1}{x}}}{x^2} \approx \infty - \frac{1}{\infty} = \infty - 0 \to \infty\). So, for large \(x\), \(g'(x) > 0\).

Since \(g'(x)\) is negative for some values in \((0, \infty)\) (near 0) and positive for other values (large x), \(g'(x)\) is not strictly less than 0 for all \(x \in (0, \infty)\).

Therefore, \(g(x)\) is not decreasing in the entire interval \((0, \infty)\). It decreases in some subinterval near 0 and increases in another subinterval towards infinity.

Thus, Statement 2 is incorrect.

Conclusion

Statement 1 is correct, and Statement 2 is incorrect.

The correct option is the one stating that only Statement 1 is correct.

Revision Table: Function Monotonicity Summary

Additional Information on Monotonicity and Derivatives

The relationship between the sign of the first derivative and the monotonicity of a function is a fundamental concept in differential calculus. It helps us understand the shape of a function's graph without plotting many points.

• Strictly Increasing: If \(f'(x) > 0\) for all \(x\) in an interval, the function is strictly increasing. This means that for any \(x_1 < x_2\) in the interval, \(f(x_1) < f(x_2)\).
• Strictly Decreasing: If \(f'(x) < 0\) for all \(x\) in an interval, the function is strictly decreasing. This means that for any \(x_1 < x_2\) in the interval, \(f(x_1) > f(x_2)\).
• Increasing (Non-decreasing): If \(f'(x) \ge 0\) for all \(x\) in an interval, the function is increasing (or non-decreasing). For any \(x_1 < x_2\) in the interval, \(f(x_1) \le f(x_2)\).
• Decreasing (Non-increasing): If \(f'(x) \le 0\) for all \(x\) in an interval, the function is decreasing (or non-increasing). For any \(x_1 < x_2\) in the interval, \(f(x_1) \ge f(x_2)\).
• Critical Points: Points where \(f'(x) = 0\) or \(f'(x)\) is undefined are called critical points. These points are potential locations for local maxima or minima, and they often mark where the function changes from increasing to decreasing or vice versa.

For the function \(g(x)\), we found that \(g'(x)\) changes sign in the interval \((0, \infty)\), indicating that it is neither strictly increasing nor strictly decreasing over the entire interval, although it might be monotonic on subintervals.