Problem 22 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=\frac{x}{x-1} $$

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x}{x-1}\) is decreasing on the intervals \((-\infty, 1)\) and \((1, \infty)\). There are no relative maxima or minima in the function.

Step 1: Find the derivative of the function

To differentiate f(x) = \(\frac{x}{x - 1}\), we can use the quotient rule. If u(x) = x and v(x) = x - 1, the quotient rule states: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} $$ Now, let's find the derivative of f(x).

Step 2: Calculating the derivative using the quotient rule

Calculate the derivatives of u(x) and v(x) and plug them into the formula to find f'(x): u'(x) = \(\frac{d}{dx}\) (x) = 1 v'(x) = \(\frac{d}{dx}\) (x - 1) = 1 Now, plug these derivatives and u(x), v(x) into the formula: $$ f'(x) = \frac{1(x - 1) - x(1)}{(x - 1)^2} = \frac{x - 1 - x}{(x - 1)^2} = \frac{-1}{(x - 1)^2} $$

Step 3: Determine the intervals on which f is increasing or decreasing

With the derivative found, we can now determine where the function is increasing and decreasing. Since the denominator \((x - 1)^2\) is always positive, with the exception of x = 1 when it is undefined, f'(x) changes its sign based on the numerator (-1) only: f'(x) < 0 for all x ≠ 1 In conclusion, f(x) is decreasing on the intervals \((-\infty, 1)\) and \((1, \infty)\), and the function is not increasing on any interval.

Step 4: Identify the relative maxima and minima

Since f(x) is decreasing on all intervals for x ≠ 1, there are no relative maxima or minima in this function. The function is continuous and strictly decreasing for all x values except 1.

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus for finding the derivative of a function that is the quotient of two other functions. This rule comes in handy when dealing with a function expressed as the division of one function by another, like \( f(x) = \frac{u(x)}{v(x)} \). In such cases, the derivative \( f'(x) \) can be calculated using the formula:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\]

• \( u(x) \) and \( v(x) \) are the numerator and denominator functions, respectively.
• \( u'(x) \) and \( v'(x) \) are the derivatives of these functions.
• The denominator in the formula, \( v(x)^2 \), ensures the function remains valid except where \( v(x) = 0 \), so discontinuities must be considered.

Using this rule, we can reliably find the derivative of complex fractions in calculus problems.

Derivative

Derivatives provide a way to understand the rate of change of functions. In a practical sense, derivatives help identify how a function's output value changes concerning its input. When we derive \( f(x) = \frac{x}{x-1} \), the quotient rule gives us:\[f'(x) = \frac{-1}{(x-1)^2}\]This tells us several important things:

• The derivative is negative, indicating that the function is decreasing.
• It is undefined when \( x = 1 \), where the original function is also undefined.

By examining the sign of \( f'(x) \) across different intervals, you can determine the behavior of the function across its domain.

Increasing and Decreasing Intervals

Determining where a function is increasing or decreasing is essential for understanding its overall behavior. We achieve this by analyzing the sign of the derivative.

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

For the function \( f(x) = \frac{x}{x-1} \), we've already calculated that \( f'(x) = \frac{-1}{(x-1)^2} \), which is negative for all \( x eq 1 \). Thus, the function is decreasing on both intervals \((-\infty, 1)\) and \((1, \infty)\). This means there are no intervals where the function is increasing.

Relative Maxima and Minima

Finding relative maxima and minima involves examining where a function changes its increasing or decreasing behavior. These points are crucial in optimization problems and understanding the function's graph shape. Generally:

• Relative maxima occur where the function shifts from increasing to decreasing.
• Relative minima occur where the function shifts from decreasing to increasing.

In our case, \( f(x) = \frac{x}{x-1} \) is strictly decreasing, indicating no change in behavior that would create relative extrema. This implies that there are no points of relative maxima or minima in the defined intervals. As a result, the function consistently trends downward, reaffirming its decreasing nature.