Problem 3 - Contemporary Precalculus

Question

Find the coordinates of the highest or lowest point on the part of the graph of the equation in the given viewing window. Only the range of \(x\) -coordinates for the window are given \(_{i}\) you must choose an appropriate range of \(y\) -coordinates. $$y=\frac{4}{x^{2}}-\frac{7}{x}+1 ; \quad \text { lowest point when }-10 \leq x \leq 10$$

Step-by-Step Solution

Answer

Answer: The lowest point of the function within the given range is at the coordinates \(\left(\frac{7}{8}, -\frac{57}{7}\right)\).

Step 1: Calculating the derivative

We begin by finding the first derivative of the given function with respect to \(x\): $$y'=\frac{d}{dx} \left(\frac{4}{x^{2}}-\frac{7}{x}+1\right)$$ Using the quotient rule, we get: $$y' = \frac{-8x}{x^4} + \frac{7}{x^2}$$

Step 2: Setting the derivative equal to zero and solving for \(x\)

To find the critical points, we set the first derivative equal to zero and solve for \(x\): $$0 = \frac{-8x}{x^4} + \frac{7}{x^2}$$ Multiply both sides by \(x^4\) to simplify the equation: $$0 = -8x^3 + 7x^2$$ Factor out \(x^2\): $$0 = x^2(-8x + 7)$$

Step 3: Analyzing critical points

From the above equation, we have 2 critical points: \(x = 0\) and \(x = \frac{7}{8}\). However, since the domain of the function is \(\{-10 \leq x \leq 10\}\), we should only consider the critical points within this interval. Therefore, our valid critical point is \(x = \frac{7}{8}\).

Step 4: Finding the lowest value of \(y\)

To find the lowest value of \(y\), we substitute the valid critical point (\(x = \frac{7}{8}\)) into the original equation: $$y = \frac{4}{\left(\frac{7}{8}\right)^{2}}-\frac{7}{\frac{7}{8}}+1$$ $$y = -\frac{57}{7}$$ Now we have the coordinates of the lowest point: \(\left(\frac{7}{8}, -\frac{57}{7}\right)\).

Step 5: Determine the appropriate range of \(y\)-coordinates

Since the lowest point of the function is at \(y = -\frac{57}{7}\), we can choose an appropriate range of \(y\)-coordinates by taking this value as the lowest value of \(y\). An appropriate range of \(y\)-coordinates could be \(-30 \leq y \leq 30\), which should cover the given range of \(x\)-coordinates, \(-10 \leq x \leq 10\).

Key Concepts

Critical points

In calculus, critical points play a vital role in understanding a function's behavior. A critical point of a function is where its derivative is either zero or undefined. These points are crucial because they often indicate local maxima, minima, or saddle points.

To find the critical points of a function, follow these steps:

• First, compute the first derivative of the function.
• Set the derivative equal to zero and solve for the variable, usually represented as \(x\).
• Critical points may also occur where the derivative does not exist.
• Finally, confirm that these critical points lie within the given domain of the function.

By analyzing critical points, you can determine the essential characteristics of the graph, such as where it attains its highest or lowest points within a particular interval.

First derivative

The first derivative of a function provides critical insights into its behavior and shapes. It measures the rate at which the function's value changes as the input changes.

The first derivative is found by differentiating the function with respect to the independent variable, often using rules like the power rule, product rule, and quotient rule.

This derivative helps ascertain:

• Where the function is increasing or decreasing.
• The critical points and possibly the points of inflection.
• The slope of the tangent at any given point.

Understanding the first derivative is fundamental in calculus, as it provides the groundwork for sketching graphs and solving problems related to rates of change.

Quotient rule

The quotient rule is a method in calculus used to find the derivative of a function that is the ratio of two differentiable functions. When you have a function \(y = \frac{u}{v}\), where both \(u\) and \(v\) depend on \(x\), the derivative is given by:

\(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}\).

This formula results from applying the chain and product rules together and is beneficial in finding derivatives of fractions with variable expressions in both numerator and denominator.

In our original exercise, the quotient rule was applied to differentiate the function \(y = \frac{4}{x^2} - \frac{7}{x} + 1\), simplifying the process of finding the first derivative.

Function graph

Understanding the graph of a function involves knowing its critical points, intercepts, regions of increase and decrease, and any asymptotic behaviors.

To sketch a function graph efficiently, follow these steps:

• Locate critical points where the derivative equals zero or is undefined. These are potential extrema.
• Analyze the sign of the first derivative to determine intervals of increasing or decreasing behavior.
• Identify the y-intercept by setting \(x = 0\) in the function, if defined.
• Look for any asymptotes by observing the limits as \(x\) approaches critical values.

The graph of a function such as \(y = \frac{4}{x^2} - \frac{7}{x} + 1\) requires considering these aspects to understand its overall shape and find significant features like the lowest point within the given interval. Bringing these insights together allows for a comprehensive picture of the function's graph.