Problem 297 - Calculus Volume 1

Question

Draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior. . \(y=\frac{x^{3}+4 x^{2}+3 x}{3 x+9}\)

Step-by-Step Solution

Answer

The function has a vertical asymptote at \( x = -3 \), a local minimum at \( x = -\frac{1}{2} \), no inflection points, and as \( x \to \infty \), it approaches \( y = \frac{x}{3} \).

Step 1: Break Down the Function

Given function is \( y = \frac{x^3 + 4x^2 + 3x}{3x + 9} \). We'll analyze the numerator and the denominator separately to understand the behavior of the graph.

Step 2: Simplify the Function

Factor both the numerator and denominator: - Numerator: \( x(x^2 + 4x + 3) \) factors to \( x(x+1)(x+3) \).- Denominator: \( 3(x+3) \). Thus the function simplifies to \( y = \frac{x(x+1)(x+3)}{3(x+3)} \).

Step 3: Identify Domain and Remove Cancellation

The term \((x+3)\) cancels out in the numerator and denominator. So the simplified function is \( y = \frac{x(x+1)}{3} \) when \( x eq -3 \). At \( x = -3 \), the function is undefined.

Step 4: Determine Asymptotic Behavior

There is a vertical asymptote at \( x = -3 \) since the original function is undefined there.For horizontal asymptotic behavior, as \( x \to \infty \) or \( x \to -\infty \), the dominant term in the numerator is \( x^2 \) and \( x \) in the denominator, giving the leading behavior \( y \sim \frac{x^2}{3x} = \frac{x}{3} \).

Step 5: Calculate Derivatives for Critical Points

Find the first derivative to identify critical points: \( f'(x) = \frac{d}{dx} \left( \frac{x^2 + x}{3} \right) = \frac{2x + 1}{3} \).Set \( f'(x) = 0 \), solving \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \). This is a critical point.

Step 6: Analyze Sign of First Derivative

Check the behavior of \( f'(x) \) around critical points:- For \( x < -\frac{1}{2} \), \( f'(x) < 0 \) (decreasing).- For \( x > -\frac{1}{2} \), \( f'(x) > 0 \) (increasing).Thus, \( x = -\frac{1}{2} \) is a local minimum.

Step 7: Second Derivative for Inflection Points

Calculate the second derivative for inflection points: \( f''(x) = \frac{d}{dx}(\frac{2x + 1}{3}) = \frac{2}{3} \).Since this is constant, there are no inflection points.

Step 8: Sketch the Graph

- The graph has a vertical asymptote at \( x = -3 \).- There's a local minimum at \( x = -\frac{1}{2} \).- The asymptote affects the graph drastically as \( x \to -3^+ \) and \( x \to -3^- \).- The function approaches a line with a slope of \( \frac{1}{3} \) as \( x \to \pm \infty \).- No inflection points due to constant second derivative.

Key Concepts

Critical Points

A critical point of a function occurs where the derivative is zero or undefined. These points are essential because they often indicate where a function might change direction, which could lead to a local maximum or minimum.
To find the critical points, we calculate the first derivative of the function. For our simplified function, this is:\[ f'(x) = \frac{2x + 1}{3}. \]

• Set the first derivative to zero: \(2x + 1 = 0\).
• Solve for \(x\) to get the critical point: \(x = -\frac{1}{2}\).

This critical point indicates a potential change in the direction of the graph. Once found, you should test points around \(x = -\frac{1}{2}\) to determine if the function is increasing or decreasing at this point, thus confirming if it is a local maximum or minimum.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches certain values or tends towards infinity.
The original function is undefined at \(x = -3\), creating a vertical asymptote. This means it stretches to infinity in this region, causing the graph to appear as though it's approaching a vertical line.

• Vertical asymptote at \(x = -3\): the graph does not touch or cross this line.
• Horizontal behavior as \(x\to \infty\): Dominant term analysis gives \(y\approx \frac{x}{3}\), indicating a slope of \(\frac{1}{3}\).

This information is crucial in helping us sketch the graph accurately, showing how it behaves near these boundaries.

Local Maxima and Minima

Local maxima and minima are points where the function reaches a small peak or trough in its value. These are indicated by critical points where the graph shifts direction.For our function, by evaluating the sign changes of the first derivative \(f'(x)\), we find:

• For \(x = -\frac{1}{2}\): It's a local minimum.
• When \(x < -\frac{1}{2}\): The function decreases.
• When \(x > -\frac{1}{2}\): The function increases.

These observations tell us that at \(x = -\frac{1}{2}\), the graph bottoms out before increasing again, forming a local minimum, but no local maximum is present in the simplified region.

Inflection Points

An inflection point is where the function changes its concavity, from smiling to frowning or vice versa. Identifying these points involves second derivatives.
We calculate the second derivative:\[ f''(x) = \frac{2}{3}. \]

• Here, \(f''(x)\) is constant, meaning it does not change sign.
• No points of inflection are present since there's no change in concavity.

Since \(f''(x)\) is constant and does not equal zero, the function maintains consistent curvature without inflection points. This information reveals more about the smooth transition between increasing and decreasing regions.