Problem 239 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. \(f(x)=\frac{1}{4} \sqrt{x}+\frac{1}{x}, x>0\)

Step-by-Step Solution

Answer

Increasing: \((0, 64)\); Decreasing: \((64, \infty)\). Local max at \(x=64\). Concave up: \((0,16)\); Concave down: \((16,\infty)\) with inflection at \(x=16\).

Step 1: Find the Derivative

First, we need to find the derivative of the function to determine where it is increasing or decreasing. We have \( f(x) = \frac{1}{4} \sqrt{x} + \frac{1}{x} \). Differentiating term by term gives us \( f'(x) = \frac{1}{8\sqrt{x}} - \frac{1}{x^2} \).

Step 2: Determine Critical Points

To find intervals of increase or decrease, solve \( f'(x) = 0 \). Set \( \frac{1}{8\sqrt{x}} = \frac{1}{x^2} \). Solving this equation, we get \( \sqrt{x} = 8 \), giving \( x = 64 \). Thus, \( x = 64 \) is a critical point. Additionally, \( x = 0 \) is considered because it is the endpoint of the domain, although \( f(x) \) is not defined at \( x = 0 \).

Step 3: Test Intervals for Increasing/Decreasing

Choose test points from intervals \((0, 64)\) and \((64, \infty)\) and plug them into \( f'(x) \) to determine the sign. For \( x < 64 \), such as \( x = 1 \), \( f'(1) \) is positive. For \( x > 64 \), such as \( x = 100 \), \( f'(100) \) is negative. Therefore, \( f(x) \) is increasing on \((0, 64)\) and decreasing on \((64, \infty)\).

Step 4: Determine Local Extrema

Since \( f(x) \) is increasing on \((0, 64)\) and decreasing on \((64, \infty)\), \( x = 64 \) is a point of local maximum.

Step 5: Find Second Derivative

To determine the concavity, find the second derivative. Differentiate \( f'(x) \) to get \( f''(x) = -\frac{1}{16x^{3/2}} + \frac{2}{x^3} \).

Step 6: Concavity and Inflection Points

Find where \( f''(x) = 0 \). Set \( -\frac{1}{16x^{3/2}} + \frac{2}{x^3} = 0 \), solve to get \( x = 16 \). For \( x < 16 \), test points show \( f''(x) > 0 \) (concave up) and for \( x > 16 \), \( f''(x) < 0 \) (concave down). Thus, \((0,16)\) is concave up and \((16, \infty)\) is concave down with an inflection point at \( x = 16 \).

Step 7: Sketch the Curve

Plot the points and use the intervals and concavity to sketch the curve. Check using a calculator to ensure accuracy; the critical point, local maximum, and inflection points should appear correctly.

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus as they help us understand how functions change. They measure the rate of change or slope of a function at any given point. To determine whether a function is increasing or decreasing, we look at the sign of its derivative. If the derivative is positive over an interval, the function is increasing; if negative, the function is decreasing.

For the function in our exercise, after differentiating, the derivative is given by \( f'(x) = \frac{1}{8\sqrt{x}} - \frac{1}{x^2} \). We solve for intervals where this is positive or negative to identify regions of growth or shrinkage. Finding points where the derivative is zero or undefined helps locate critical points, pivotal for determining local extrema.

Exploring Local Extrema

Local extrema refer to points where a function reaches a local maximum or minimum value. To find these points, we analyze the critical points of the derivative, where \(f'(x) = 0\) or is undefined. By evaluating how the function behaves around these points, we can classify them as local maxima or minima.

In our problem, the critical point occurs at \(x = 64\). As the function increases before and decreases after this point, \(x = 64\) is a local maximum. Recognizing local extrema is crucial as they often highlight important features of the function, indicating potential turning points in its graph.

Understanding Concavity

Concavity describes how a function curves. If a function's graph is shaped like a bowl facing upwards, it is concave up; if it resembles a bowl facing downwards, it's concave down. The second derivative, \(f''(x)\), tells us about concavity. If \(f''(x) > 0\), the function is concave up; if \(f''(x) < 0\), it is concave down.

For the given function, \(f''(x) = -\frac{1}{16x^{3/2}} + \frac{2}{x^3}\). By examining this, we find the function is concave up over \((0, 16)\) and concave down over \((16, \infty)\). These intervals help in sketching the curve accurately, providing insight into the behavior of the function across its domain.

Identifying Inflection Points

Inflection points occur where the graph of a function changes its concavity from upwards to downwards or vice versa. At these points, the second derivative is zero or undefined. They are significant as they mark a change in the direction of curvature, often indicating a shift in the behavior of the function.

In the exercise, the inflection point was identified by setting the second derivative to zero, leading to the solution \(x = 16\). This inflection point is where the function shifts from concave up to concave down. Detecting inflection points is essential in plotting a complete and precise graph of the function, offering a deeper view of its characteristics.