Problem 65 - Calculus: Early Transcendentals

Question

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=4 x+\sqrt{1-x} $$

Step-by-Step Solution

Answer

The curve is concave up, has a y-intercept at (0,1), no x-intercept, and a critical point near \( x=\frac{63}{64} \).

Step 1: Identify Domain

Since the function involves a square root \( \sqrt{1-x} \), we need to ensure that the expression inside the square root is non-negative. Thus, \( 1-x \geq 0 \) leads to \( x \leq 1 \). Therefore, the domain of the function is \( x \in (-\infty, 1] \).

Step 2: Find Intercepts

To find the y-intercept, set \( x = 0 \). So, \( y = 4(0) + \sqrt{1-0} = 1 \). Thus, the y-intercept is (0, 1). To find the x-intercept, set \( y = 0 \), giving the equation \( 4x + \sqrt{1-x} = 0 \). On solving, this doesn't yield a simple analytical solution, so there is no x-intercept.

Step 3: Analyze Asymptotic Behavior

The function \( y = 4x + \sqrt{1-x} \) doesn't have any vertical asymptotes since it's defined for \( x \leq 1 \). As \( x \to -\infty \), \( 4x \) dominates and \( y \to -\infty \). There is no horizontal asymptote.

Step 4: Calculate Derivative for Critical Points

Finding the derivative \( y' = 4 - \frac{1}{2\sqrt{1-x}} \). Set this to zero to determine critical points: \( 4 - \frac{1}{2\sqrt{1-x}} = 0 \). Simplifying, \( \sqrt{1-x} = \frac{1}{8} \), so \( 1-x = \frac{1}{64} \), and hence \( x = \frac{63}{64} \). Evaluate \( y \) at this point for a critical point value.

Step 5: Analyze Second Derivative for Concavity and Inflection Points

Calculate the second derivative \( y'' = \frac{1}{4(1-x)^{3/2}} \). For \( x < 1 \), \( y'' > 0 \), meaning the curve is concave up. Thus, no inflection points exist as the concavity does not change.

Step 6: Plot Important Features and Sketch the Curve

Plot the y-intercept (0, 1) and the critical point \( \left(\frac{63}{64}, 4\frac{63}{64} + \sqrt{1-\frac{63}{64}} \right) \). The function is concave up for the entire domain. Sketch an increasing curve from negative infinity, rising steeply and becoming smoother near the point (1, 4).

Key Concepts

Domain of a function

When determining the domain of a function, especially one involving square roots like \( y=4x+\sqrt{1-x} \), it is crucial to focus on the expression inside the square root. This is because square roots of negative numbers are not defined within the real number system. Here, for the square root \( \sqrt{1-x} \) to be real, the expression \( 1-x \) must be non-negative.
Therefore, the inequality \( 1-x \geq 0 \) is solved as follows:

• Rearranging gives \( x \leq 1 \)
• The domain thus becomes all real numbers \( x \) such that \( x \leq 1 \)

This domain restriction means that the function is defined from negative infinity up to and including 1. So, \( x \in (-\infty, 1] \). Each function's domain can deeply affect its graph and behavior, making this an essential first step in understanding any function.

Critical points

Critical points are where the derivative of a function is zero or undefined. For the function \( y=4x+\sqrt{1-x} \), we first need the first derivative \( y' \) to find these points.
Calculating the derivative:

• The derivative is \( y' = 4 - \frac{1}{2\sqrt{1-x}} \)
• Setting \( y' = 0 \) gives the equation \( 4 - \frac{1}{2\sqrt{1-x}} = 0 \)

Solving, we end up with \( \sqrt{1-x} = \frac{1}{8} \), leading to \( x = \frac{63}{64} \). At \( x = \frac{63}{64} \), the derivative equals zero, indicating a possible local extremum. Calculating \( y \) at this point verifies it as a critical value.
Identifying critical points helps find where a function might change from increasing to decreasing or vice versa, revealing potential local maximums or minimums.

Concavity and inflection points

The concavity of a function determines whether it curves upwards or downwards. To assess concavity, we look at the second derivative. For \( y=4x+\sqrt{1-x} \), the second derivative \( y'' \) is calculated as follows:

• \( y'' = \frac{1}{4(1-x)^{3/2}} \)

For \( x < 1 \), \( y'' > 0 \), which means the function is concave up for the entire domain. A function that is always concave up will never have an inflection point within the given domain because the concavity does not change sign. Inflection points occur where a function changes from concave up to concave down or vice versa, but such a change does not happen here.
Understanding a function's concavity enables us to predict its graphical behavior and highlight potential turning points.

Asymptotic behavior

Asymptotic behavior studies a function's end behavior as it approaches some limit, often infinity. In \( y=4x+\sqrt{1-x} \), no vertical asymptotes exist within the domain \( x \leq 1 \), as the function is defined and continuous. A vertical asymptote might have occurred if the function was undefined for certain \( x \) values within the domain.
Examining how \( y\) behaves as \( x \rightarrow -\infty \):

• \( y = 4x + \sqrt{1-x} \) shows \( 4x \) as the dominant term when \( x \) becomes very negative
• This results in \( y \rightarrow -\infty \)

There is no horizontal asymptote because the function does not approach a fixed \( y \) value as \( x \rightarrow -\infty \).
Understanding asymptotic behavior can often give insight into a function's long-term trends and guide the sketching of its curve.