Problem 9
Question
Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=(x+1) /(x-1) $$
Step-by-Step Solution
Verified Answer
The function is decreasing on \((-\infty, 1)\) and \((1, \infty)\) with no local maxima or minima.
1Step 1: Find the First Derivative
To determine the intervals where the function is increasing or decreasing, we first find the first derivative of the function \(f(x) = \frac{x+1}{x-1}\). We use the quotient rule for derivatives, which is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \(u = x+1\) and \(v = x-1\), so \(u' = 1\) and \(v' = 1\). Thus, \(f'(x) = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{x - 1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2}\).
2Step 2: Determine Critical Points
Critical points occur where the first derivative is zero or undefined. Here, \(f'(x) = \frac{-2}{(x-1)^2}\). Since the numerator \(-2\) is never zero, there are no points where \(f'(x) = 0\). However, \(f'(x)\) is undefined where \(x-1=0\), i.e., at \(x=1\). Thus, \(x = 1\) is a critical point where the derivative is undefined.
3Step 3: Analyze Intervals Around Critical Points
To determine where the function is increasing or decreasing, analyze the sign of \(f'(x)\) on intervals around the critical point \(x=1\). For \(x < 1\), choose a test point such as \(x=0\). Compute \(f'(0) = \frac{-2}{(0-1)^2} = -2\), which is negative, indicating the function is decreasing on \((-\infty, 1)\). For \(x > 1\), choose a test point such as \(x=2\). Compute \(f'(2) = \frac{-2}{(2-1)^2} = -2\), which is also negative, indicating the function is decreasing on \((1, \infty)\).
4Step 4: Apply the First Derivative Test
Since the derivative is negative on both intervals \((-\infty, 1)\) and \((1, \infty)\), the function is continuously decreasing and does not change direction at \(x=1\). Therefore, there is neither a local maximum nor a local minimum at \(x=1\) because \(f(x)\) does not move from increasing to decreasing or vice versa.
Key Concepts
Increasing and Decreasing IntervalsCritical PointsQuotient RuleLocal Maximum and Minimum
Increasing and Decreasing Intervals
A key part of analyzing a function involves determining where it is increasing or decreasing. A function is said to be **increasing** on an interval if, as the input values (x) get larger, the output values (f(x)) also rise. Conversely, a function is **decreasing** when an increase in x corresponds to a decrease in f(x). To find these intervals, we use the first derivative of the function to assess the slope.
For the function given, \( f(x) = \frac{x+1}{x-1} \), we calculate its first derivative using the quotient rule: \( f'(x) = \frac{-2}{(x-1)^2} \). Analyzing this derivative through test points, we find that it is negative for both \( x < 1 \) and \( x > 1 \).
Having a negative derivative indicates the function is consistently decreasing on these intervals: \(( -\infty, 1 )\) and \(( 1, \infty )\).
For the function given, \( f(x) = \frac{x+1}{x-1} \), we calculate its first derivative using the quotient rule: \( f'(x) = \frac{-2}{(x-1)^2} \). Analyzing this derivative through test points, we find that it is negative for both \( x < 1 \) and \( x > 1 \).
Having a negative derivative indicates the function is consistently decreasing on these intervals: \(( -\infty, 1 )\) and \(( 1, \infty )\).
Critical Points
Critical points are locations on the graph where the function's rate of change could switch direction; this is where the first derivative is zero or undefined. These points help identify potential peaks or troughs, known as local extrema, in the function's graph. For the function given, we look at where \( f'(x) \) is zero or undefined. In this case, \( f'(x) = \frac{-2}{(x-1)^2} \).
Because the numerator \(-2\) never equals zero, there are no points where \( f'(x) = 0 \). However, \( f'(x) \) is undefined at \( x = 1 \) (since the denominator becomes zero), marking \( x = 1 \) as a critical point.
This critical point is essential for determining any changes in the function's increasing or decreasing patterns.
Because the numerator \(-2\) never equals zero, there are no points where \( f'(x) = 0 \). However, \( f'(x) \) is undefined at \( x = 1 \) (since the denominator becomes zero), marking \( x = 1 \) as a critical point.
This critical point is essential for determining any changes in the function's increasing or decreasing patterns.
Quotient Rule
When we encounter a function involving division, such as \( f(x) = \frac{u}{v} \), computing its derivative requires the quotient rule. This rule states: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).
Here, \( u = x+1 \) and \( v = x-1 \) allow us to find \( u' = 1 \) and \( v' = 1 \). Substituting these into the rule, we get the derivative \( f'(x) = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{-2}{(x-1)^2} \).
This formula helps us understand how the slope changes at different points in the function. Using the quotient rule is vital for solving problems that involve derivatives of fractional functions, making complex five-part components manageable with structured steps.
Here, \( u = x+1 \) and \( v = x-1 \) allow us to find \( u' = 1 \) and \( v' = 1 \). Substituting these into the rule, we get the derivative \( f'(x) = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{-2}{(x-1)^2} \).
This formula helps us understand how the slope changes at different points in the function. Using the quotient rule is vital for solving problems that involve derivatives of fractional functions, making complex five-part components manageable with structured steps.
Local Maximum and Minimum
Local maxima and minima are critical, as they represent the highest and lowest points in a particular interval of a function, respectively. Although \( f'(x) \) is usually zero at these points, it's not mutually exclusive. Sometimes they occur where the derivative is undefined, like in this exercise. To identify a local max or min, apply the First Derivative Test: assessing the behavior before and after the critical points.
The function \( f(x) = \frac{x+1}{x-1} \) provides the critical point at \( x = 1 \). Reviewing the derivative, it's negative both before and after \( x = 1 \).
As the derivative does not change sign, the function doesn’t possess a local max or min at \( x = 1 \). This scenario illustrates why checking intervals around critical points is crucial for determining the ups and downs of the function, rather than just relying on when \( f'(x) = 0 \).
The function \( f(x) = \frac{x+1}{x-1} \) provides the critical point at \( x = 1 \). Reviewing the derivative, it's negative both before and after \( x = 1 \).
As the derivative does not change sign, the function doesn’t possess a local max or min at \( x = 1 \). This scenario illustrates why checking intervals around critical points is crucial for determining the ups and downs of the function, rather than just relying on when \( f'(x) = 0 \).
Other exercises in this chapter
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