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Magazine Chapter 4: Problem 31
a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=x-6 \sqrt{x-1}$$
Short Answer
Expert verified
The function decreases on (1, 10) and increases on (10, 鈭�), with a local minimum of -8 at x = 10.
Step by step solution
01
Find the Derivative
To determine where the function is increasing or decreasing, begin by computing the derivative of the function. The function is given as \( f(x) = x - 6\sqrt{x-1} \). Differentiate with respect to \( x \): \[ f'(x) = 1 - \frac{6}{2\sqrt{x-1}} = 1 - \frac{3}{\sqrt{x-1}}. \]
02
Determine Critical Points
Set the derivative \( f'(x) = 0 \) to find critical points, which may indicate intervals of increase or decrease. Solve:\[ 1 - \frac{3}{\sqrt{x-1}} = 0. \]This simplifies to \( \sqrt{x-1} = 3 \), thus \( x-1 = 9 \) and \( x = 10 \). Also note the domain restriction, so \( x > 1 \).
03
Test Intervals Around Critical Points
Identify intervals using the critical point \( x = 10 \) and the domain restriction. Check the sign of \( f'(x) \) in the intervals \((1,10)\) and \((10, \infty)\): 1. For \( x \) in \( (1, 10) \), choose \( x = 5 \): \[ f'(5) = 1 - \frac{3}{\sqrt{5-1}} = 1 - \frac{3}{2} < 0. \] This interval is decreasing.2. For \( x \) in \( (10, \infty) \), choose \( x = 11 \): \[ f'(11) = 1 - \frac{3}{\sqrt{11-1}} = 1 - \frac{3}{\sqrt{10}} > 0. \] This interval is increasing.
04
Identify Local Extrema
The critical point \( x = 10 \) could be a local extremum. Since the derivative changes sign from negative to positive, \( f(x) \) has a local minimum at \( x = 10 \). Compute the minimum value: \[ f(10) = 10 - 6\sqrt{10-1} = 10 - 18 = -8. \]There are no absolute extrema because the function values can decrease without bound as \( x \to 1^+ \).
05
Summarize Results
- The function \( f(x) = x - 6\sqrt{x-1} \) is decreasing on \((1, 10)\) and increasing on \((10, \infty)\).- There is a local minimum at \( x = 10 \) with the value \( -8 \). - The function does not have any absolute maxima or minima.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
The derivative of a function provides insight into its behavior, specifically highlighting where it increases or decreases. For our function, \( f(x) = x - 6\sqrt{x-1} \), finding the derivative is crucial.
- The derivative, denoted \( f'(x) \), represents the rate of change or the slope of the tangent line to the function at any given point.
- To differentiate \( f(x) = x - 6\sqrt{x-1} \), apply the chain rule to the square root expression to find \[ f'(x) = 1 - \frac{6}{2\sqrt{x-1}} = 1 - \frac{3}{\sqrt{x-1}}. \]
Critical Points
Critical points in calculus are where the derivative is zero or undefined. These points are important because they possibly indicate where the graph of a function changes direction.
- Set the derivative, \( f'(x) = 1 - \frac{3}{\sqrt{x-1}} \), to zero to find critical points: a necessary step to discern changes in increasing or decreasing behavior.
- Solving \[ 1 - \frac{3}{\sqrt{x-1}} = 0 \] yields \( \sqrt{x-1} = 3 \), leading to \( x = 10 \).
- Always consider domain restrictions: Since \( \sqrt{x-1} \) exists only for \( x > 1 \), ensure that our results comply with this restriction.
Increasing and Decreasing Intervals
To determine where a function is increasing or decreasing, check the sign of the derivative \( f'(x) \) around critical points. In this exercise, we've identified \( x = 10 \) as a critical point.
- Choose test points in intervals divided by critical points and domain restrictions:
- Interval \((1, 10)\): Testing with \( x = 5 \), \[ f'(5) = 1 - \frac{3}{2} = -\frac{1}{2} \] (negative), the function decreases here.
- Interval \((10, \infty)\): Testing with \( x = 11 \), \[ f'(11) = 1 - \frac{3}{\sqrt{10}} \approx 0.05 \] (positive), indicating an increasing function here.
Local Extrema
Local extrema are points where a function achieves a local minimum or maximum value. To locate these, examine points where the derivative changes sign:
- For our function, at point \( x = 10 \), \( f(x) \) could exhibit a local extremum because the sign of \( f'(x) \) changes: from negative (decreasing) to positive (increasing).
- Thus, \( x = 10 \) is a local minimum, confirmed by calculating \( f(10) = 10 - 18 = -8 \).
- No absolute extrema exists here since the function value lowers indefinitely as \( x \to 1^+ \).
