Problem 239 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The function is increasing on \((\frac{1}{64}, \infty)\), has a local minimum at \(x = \frac{1}{64}\), and the behavior of concavity changes based on derivative sign analysis.

Step 1: Identify the function and domain

The function given is \( f(x) = \frac{1}{4} \sqrt{x} + \frac{1}{x} \) with the domain \( x > 0 \). Our task is to find the intervals where the function is increasing or decreasing, its local minima and maxima, intervals of concavity, and inflection points.

Step 2: Find the first derivative

To determine the intervals where the function is increasing or decreasing, find the first derivative \( f'(x) \). Using the power rule, we find: \[ f'(x) = \frac{1}{8\sqrt{x}} - \frac{1}{x^2}. \]

Step 3: Determine critical points

Set \( f'(x) = 0 \) to find the critical points: \[ \frac{1}{8\sqrt{x}} = \frac{1}{x^2}. \] Solve for \( x \) to find the critical points. Multiply both sides by \( 8x^2\sqrt{x} \) to clear fractions, resulting in \( \sqrt{x} = 8x^{3/2} \). Solving this gives \( x = \frac{1}{64} \). Also note any points where \( f'(x) \) is undefined or changes sign.

Step 4: Test intervals for increasing/decreasing behavior

Choose test values in the intervals \( (0, \frac{1}{64}) \) and \( (\frac{1}{64}, \infty) \) to determine whether \( f'(x) \) is positive or negative. If \( f'(x) > 0 \), the function is increasing. If \( f'(x) < 0 \), the function is decreasing.

Step 5: Find the second derivative

To analyze concavity, find the second derivative \( f''(x) \). The second derivative is \[ f''(x) = -\frac{1}{16x^{3/2}} + \frac{2}{x^3}. \]

Step 6: Determine concavity and inflection points

Set \( f''(x) = 0 \) to find inflection points: \[ -\frac{1}{16x^{3/2}} + \frac{2}{x^3} = 0. \] Solve for \( x \) or analyze sign changes in \( f''(x) \). Testing intervals around potential inflection points reveals where the function is concave up (\( f''(x) > 0 \)) or concave down (\( f''(x) < 0 \)).

Step 7: Summarize findings and sketch the curve

Compile the intervals of increase/decrease, local maxima/minima, concavity, and inflection points. Use this information to sketch the graph. For verification, use a graphing calculator to compare the analytical results with the numerical plot.

Key Concepts

Function Derivatives

In calculus, a derivative helps us understand how a function behaves and changes. When we take the derivative of a function, we are essentially finding its slope at any point. This is crucial when analyzing how the function behaves across its domain.
For the given function, \( f(x) = \frac{1}{4} \sqrt{x} + \frac{1}{x} \), the first derivative \( f'(x) \) is determined using basic differentiation rules.

• The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \) because \( \sqrt{x} \) is \( x^{1/2} \) and using the power rule, we subtract 1 from the exponent giving us \( \frac{1}{2}x^{-1/2} \).
• For \( \frac{1}{x} \), which is \( x^{-1} \), the derivative is \( -x^{-2} \) or \( -\frac{1}{x^2} \).

Combining these, the first derivative is \( f'(x) = \frac{1}{8 \sqrt{x}} - \frac{1}{x^2} \). This derivative helps us identify critical points, discussed next.

Critical Points

Critical points are values of \( x \) where the function's first derivative equals zero or is undefined. These points are crucial because they might indicate local maxima or minima—places where the function changes from increasing to decreasing or vice versa.
In our problem, solving \( f'(x) = 0 \) gives us potential critical points. The equation \( \frac{1}{8 \sqrt{x}} = \frac{1}{x^2} \) simplifies by multiplying through to eliminate fractions. After simplification, we find \( x = \frac{1}{64} \) as a critical point. Additionally, we should consider where \( f'(x) \) is undefined, if applicable, but here it's defined for all \( x > 0 \).
By testing intervals around this point, we can determine if the function increases or decreases, allowing us to locate any local maxima or minima.

Concavity

Concavity tells us how the slope of the function's curve is changing. An easy way to visualize concavity is to think about how a bowl appears—concave up (") is like a smiley face, and concave down (") is like a frown.
To find concavity, we take the second derivative, \( f''(x) \). For our function, the second derivative is \( f''(x) = -\frac{1}{16x^{3/2}} + \frac{2}{x^3} \).

• When \( f''(x) > 0 \), the function is concave up. This indicates the graph is shaped like an upward-opening curve.
• When \( f''(x) < 0 \), the function is concave down, suggesting the curve opens downwards.

Analyzing the sign of \( f''(x) \) in different intervals tells us more about the concavity of different regions of the function.

Inflection Points

Inflection points are where the curve changes concavity; the graph switches from concave up to concave down or vice versa. These points are vital for accurately sketching the shape of the function.
To find inflection points, solve \( f''(x) = 0 \). For our function, the equation is \(-\frac{1}{16x^{3/2}} + \frac{2}{x^3} = 0\). Solving this will give us potential inflection points. However, it is crucial to test values around these potential points to confirm a change in concavity.
The determination of inflection points not only helps with understanding more about the function's shape but also supports the analysis of more complex functions in calculus.