Analyzing the Monotonicity of f(x) = (e^x - 1)/x

The question asks about the monotonicity of the function f(x) for the interval where x > 0. The function is defined as:

\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{\rm{x}}},{\rm{\;\;x}} > 0}\\ {0,{\rm{\;\;\;x}} = 0} \end{array}} \right.\)

To determine if a function is strictly increasing or decreasing on an interval, we need to examine the sign of its first derivative on that interval. For x > 0, the function is given by \(f(x) = \frac{e^x - 1}{x}\).

Calculating the Derivative of f(x) for x > 0

We will use the quotient rule to find the derivative of \(f(x)\) for x > 0. The quotient rule states that if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\).

• Let \(u(x) = e^x - 1\). Then \(u'(x) = \frac{d}{dx}(e^x - 1) = e^x\).
• Let \(v(x) = x\). Then \(v'(x) = \frac{d}{dx}(x) = 1\).

Now, applying the quotient rule:

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

\(f'(x) = \frac{xe^x - e^x + 1}{x^2}\)

Determining the Sign of f'(x) for x > 0

To determine the monotonicity of f(x) for x > 0, we need to find the sign of \(f'(x)\) in this interval.

The denominator of \(f'(x)\) is \(x^2\). For x > 0, \(x^2\) is always positive.

We need to analyze the sign of the numerator, which is \(g(x) = xe^x - e^x + 1\). Let's examine the derivative of \(g(x)\):

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

Using the product rule for \(xe^x\) (\(\frac{d}{dx}(uv) = u'v + uv'\)):

\(g'(x) = (1 \cdot e^x + x \cdot e^x) - e^x + 0\)

\(g'(x) = e^x + xe^x - e^x\)

\(g'(x) = xe^x\)

For x > 0, \(x\) is positive and \(e^x\) is positive. Therefore, \(g'(x) = xe^x\) is positive for all x > 0.

Since \(g'(x) > 0\) for all x > 0, the function \(g(x) = xe^x - e^x + 1\) is strictly increasing for all x > 0.

Now let's consider the limit of \(g(x)\) as x approaches 0 from the right:

\(\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (xe^x - e^x + 1) = (0 \cdot e^0 - e^0 + 1) = 0 - 1 + 1 = 0\)

Since \(g(x)\) is strictly increasing for x > 0 and approaches 0 as x approaches 0 from the right, this means that for all x > 0, \(g(x) > 0\). That is, \(xe^x - e^x + 1 > 0\) for x > 0.

Therefore, the numerator of \(f'(x)\) is positive for x > 0, and the denominator \(x^2\) is also positive for x > 0.

\(f'(x) = \frac{\text{positive}}{\text{positive}} > 0\)

So, \(f'(x) > 0\) for all x > 0.

Interpreting the Derivative and Monotonicity

If the first derivative of a function is positive over an interval, the function is strictly increasing over that interval. Since we found that \(f'(x) > 0\) for all x > 0, the function f(x) is strictly increasing in the interval (0, \(\infty\)). This implies that f(x) is strictly increasing in any sub-interval (0, x) where x > 0.

Evaluating the Options

Let's examine the given options based on our findings:

• Option 1: f(x) is a strictly decreasing function in (0, x). This is incorrect because \(f'(x) > 0\) for x > 0.
• Option 2: f(x) is a strictly increasing function in (0, x). This is correct because \(f'(x) > 0\) for x > 0.
• Option 3: f(x) is neither increasing nor decreasing in (0, x). This is incorrect because \(f'(x)\) has a consistent positive sign in (0, x).
• Option 4: f(x) is not decreasing in (0, x). This statement is true if the function is increasing or constant. Since we found it is strictly increasing, it is indeed not decreasing. However, Option 2 provides a more precise description of the function's behavior. In multiple-choice questions, the most accurate and specific correct answer is typically required. "Strictly increasing" is a more specific and accurate description than simply "not decreasing".

Summary of Findings

Based on the derivative test, \(f'(x) = \frac{xe^x - e^x + 1}{x^2}\). By analyzing the sign of the numerator and denominator, we determined that \(f'(x) > 0\) for all x > 0. This conclusively shows that the function f(x) is strictly increasing in the interval (0, x) for any x > 0.

Revision Table: Key Concepts in Monotonicity

Additional Information: Taylor Series for e^x and f(x) Monotonicity

Another way to understand why \(f'(x) > 0\) for x > 0 is by using the Taylor series expansion of \(e^x\) around x=0:

\(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\) for all real x.

For x > 0, we can write:

\(e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\)

Dividing by x (since x > 0, we don't divide by zero):

\(f(x) = \frac{e^x - 1}{x} = \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...}{x}\)

\(f(x) = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + ...\) for x > 0

Now, let's differentiate this series term by term for x > 0:

\(f'(x) = \frac{d}{dx} \left( 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + ... \right)\)

\(f'(x) = 0 + \frac{1}{2!} + \frac{2x}{3!} + \frac{3x^2}{4!} + \frac{4x^3}{5!} + ...\)

\(f'(x) = \frac{1}{2} + \frac{2x}{6} + \frac{3x^2}{24} + \frac{4x^3}{120} + ...\)

\(f'(x) = \frac{1}{2} + \frac{x}{3} + \frac{x^2}{8} + \frac{x^3}{30} + ...\)

For x > 0, every term in this series is positive (\(\frac{1}{2}\) is positive, \(\frac{x}{3}\) is positive, \(\frac{x^2}{8}\) is positive, and so on). Therefore, the sum of all terms is positive.

\(f'(x) > 0\) for all x > 0.

This confirms that the function f(x) is strictly increasing for x > 0, aligning with the result obtained using the quotient rule on the closed form of the function.