Question

Determine the values of \(x\) for which \[ f(x)=\frac{x-3}{x+1}, \quad x\neq -1 \] is an increasing function. 

Hint:

If the derivative of a rational function simplifies to a positive constant divided by a squared term, it is strictly increasing in its entire domain.
Always state excluded domain points from the original function.

Answer 1

Concept:
• A function \(f(x)\) is increasing in an interval if its first derivative \(f'(x) > 0\).
• Quotent Rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\).

Step 1: Differentiate \(f(x)\) with respect to \(x\)
Using the quotient rule where \(u = x - 3\) and \(v = x + 1\): \[ f'(x) = \frac{(x + 1) \cdot \frac{d}{dx}(x - 3) - (x - 3) \cdot \frac{d}{dx}(x + 1)}{(x + 1)^2} \] \[ f'(x) = \frac{(x + 1)(1) - (x - 3)(1)}{(x + 1)^2} \]

Step 2: Simplify the derivative expression
\[ f'(x) = \frac{x + 1 - x + 3}{(x + 1)^2} \] \[ f'(x) = \frac{4}{(x + 1)^2} \]

Step 3: Analyze the sign of the derivative
For \(f(x)\) to be increasing, we need \(f'(x) > 0\). The numerator is \(4\), which is always positive. The denominator is \((x + 1)^2\), which is a perfect square and is always positive for all \(x \neq -1\). Since both numerator and denominator are positive, \(f'(x) > 0\) for all \(x \in \mathbb{R}\) except \(x = -1\).

Step 4: State the final interval
The function is increasing for: \[ x \in (-\infty, -1) \cup (-1, \infty) \] Or simply, \(x \in \mathbb{R} - \{-1\}\).