Problem 60 - Essential Calculus Early Transcendentals

Question

Produce graphs of \(f\) that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly. $$f(x)=\frac{1}{x^{8}}-\frac{2 \times 10^{8}}{x^{4}}$$

Step-by-Step Solution

Answer

The function increases on \(( -\frac{1}{5000}, \frac{1}{5000}) \) and is concave up on \(( -\infty, -\frac{1}{2500}) \cup (\frac{1}{2500}, \infty) \).

Step 1: Find the First Derivative

To find the intervals of increase and decrease, we first need the first derivative of the function. The function is \( f(x) = \frac{1}{x^8} - \frac{2 \times 10^8}{x^4} \). We differentiate it using the power rule: \[ f'(x) = -8x^{-9} + 4 \times 10^8 x^{-5}. \]

Step 2: Solve for Critical Points

Set the first derivative to zero to find critical points: \[ -8x^{-9} + 4 \times 10^8 x^{-5} = 0. \] Simplifying, we get \( -8 + 4 \times 10^8 x^4 = 0 \).Therefore, \( x^4 = \frac{8}{4 \times 10^8} \).Solving gives \( x = \pm \frac{1}{5000} \). These are the critical points.

Step 3: Test Intervals for Increase and Decrease

Test the sign of \( f'(x) \) in the intervals determined by the critical points. Choose test points in each interval: when \( x < -\frac{1}{5000} \), \( -\frac{1}{5000} < x < \frac{1}{5000} \), and \( x > \frac{1}{5000} \). This will help determine if the function is increasing or decreasing in each interval.

Step 4: Find the Second Derivative

To find concavity, compute the second derivative of the function: \[ f''(x) = 72x^{-10} - 20 \times 10^8 x^{-6}. \]

Step 5: Solve for Points of Inflection

Set the second derivative equal to zero: \[ 72x^{-10} - 20 \times 10^8 x^{-6} = 0. \] Simplifying gives \( x^4 = \frac{72}{20 \times 10^8} \). Solving results in potential inflection points at \( x = \pm \frac{1}{2500} \).

Step 6: Test Intervals for Concavity

Test the sign of \( f''(x) \) around the potential points of inflection to determine concave up or down in the intervals \( (-\infty, -\frac{1}{2500}) \), \( (-\frac{1}{2500}, \frac{1}{2500}) \), and \( (\frac{1}{2500}, \infty) \).

Step 7: Sketching the Graph

Based on the analysis from previous steps, sketch the graph showing the intervals of increase, decrease, and concavity. Highlight the critical points and inflection points determined.

Key Concepts

Derivative Calculation

Calculating derivatives is a fundamental skill in calculus that helps us understand how functions change. Derivatives are used to determine the rate of change of a function at any given point. For the function in this problem,

• we use the power rule to find the first derivative.

Given the function: \[ f(x) = \frac{1}{x^8} - \frac{2 \times 10^8}{x^4} \] We differentiate it:
\[ f'(x) = -8x^{-9} + 4 \times 10^8 x^{-5} \] The power rule involves multiplying by the exponent and reducing the exponent by one:

• \( x^{-8} \) becomes \(-8x^{-9} \)
• \( x^{-4} \) becomes \(-4 \times 10^8 x^{-5} \)

Once you discover the first derivative, it reveals where the function is either increasing or decreasing depending on the sign of the derivative across different intervals of the domain. Differentiation and these calculations allow us to determine critical points, which are essential for further analysis.

Critical Points

Critical points are where a function's derivative is zero or undefined. They help identify where there might be a change of direction in the function, i.e., from increasing to decreasing or vice versa.
For this exercise:

• We set the first derivative to zero to solve for the critical points:

\[ -8x^{-9} + 4 \times 10^8 x^{-5} = 0 \]This simplifies to \[ -8 + 4 \times 10^8 x^4 = 0 \]Solving gives \[ x^4 = \frac{8}{4 \times 10^8} \]which further simplifies to \[ x = \pm \frac{1}{5000} \].
Once found, these critical points divide the number line into intervals. By choosing test points in each interval, you can evaluate the function's behavior to determine whether it is increasing or decreasing in those sections. Recognizing these intervals assists with graph analysis, predicting potential peaks and troughs in the curve behavior.

Concavity and Inflection Points

Concavity describes how a function curves. It is measured by the second derivative, indicating whether a function is curving upwards (concave up) or downwards (concave down).
Calculate the second derivative of the function:

• Given: \[ f''(x) = 72x^{-10} - 20 \times 10^8 x^{-6} \]

To find inflection points, set the second derivative to zero:

• \[ 72x^{-10} - 20 \times 10^8 x^{-6} = 0 \]
• Solving gives: \[ x^4 = \frac{72}{20 \times 10^8} \] which results in potential inflection points \( x = \pm \frac{1}{2500} \)

Inflection points are where a function changes concavity. Test intervals around these points by plugging into the second derivative to confirm the concavity change.
These insights guide the sketching of the curve, enhancing our understanding of the function's true nature. Concave up suggests a mimicking of a U-shape; whereas, concave down looks more like an ∩-shape. Understanding this concept is paramount for accurate graph depiction.