Curve Sketching: Videos & Practice Problems

Understanding a function's first and second derivatives is crucial for graphing. The first derivative indicates where a function is increasing or decreasing, while the second derivative reveals concavity. For the function f(x) = x³ - 3x² + 4, we find critical points, x and y intercepts, and analyze concavity to identify local maxima and minima. This process involves determining intervals of increase and decrease, identifying inflection points, and sketching the graph based on these insights, ultimately leading to a comprehensive understanding of the function's behavior.

Summary of Curve Sketching

Understanding the behavior of a function through its first and second derivatives is crucial for sketching its graph. The first derivative provides insights into where the function is increasing or decreasing, while the second derivative reveals information about concavity. This chapter focuses on applying these concepts to graph the function \( f(x) = x^3 - 3x^2 + 4 \).

First, we determine the domain of the function. As a polynomial, its domain is all real numbers, expressed as \( (-\infty, \infty) \). Next, we find the x-intercepts by setting the function equal to zero. Factoring the polynomial gives us \( (x - 2)^2(x + 1) \). Setting each factor to zero results in x-intercepts at \( x = 2 \) and \( x = -1 \). The y-intercept is found by evaluating \( f(0) \), yielding \( f(0) = 4 \), so the y-intercept is at \( (0, 4) \).

Next, we check for asymptotes. Since this is a polynomial function, there are no asymptotes to consider. To analyze symmetry, we substitute \( -x \) into the function. The resulting expression does not match the original function, indicating that it is neither symmetric about the y-axis nor the origin.

Moving on to the first derivative, we differentiate \( f(x) \) to find \( f'(x) = 3x^2 - 6x \), which factors to \( 3x(x - 2) \). Setting the first derivative to zero gives critical points at \( x = 0 \) and \( x = 2 \). We create a sign chart to analyze the intervals: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \). Testing values from each interval reveals that the function is increasing on \( (-\infty, 0) \) and \( (2, \infty) \), and decreasing on \( (0, 2) \).

Next, we examine concavity using the second derivative. Differentiating \( f'(x) \) gives \( f''(x) = 6x - 6 \), which factors to \( 6(x - 1) \). Setting this equal to zero identifies a potential inflection point at \( x = 1 \). Testing intervals \( (-\infty, 1) \) and \( (1, \infty) \) shows that the function is concave down on \( (-\infty, 1) \) and concave up on \( (1, \infty) \).

With this information, we can identify local extrema. At \( x = 0 \), the function transitions from increasing to decreasing, indicating a local maximum. At \( x = 2 \), the function transitions from decreasing to increasing, indicating a local minimum. Evaluating these points in the original function confirms their coordinates: \( (0, 4) \) for the local maximum and \( (2, 0) \) for the local minimum.

Finally, we sketch the graph. Starting from the left, the function increases and is concave down until reaching the local maximum at \( (0, 4) \). It then decreases while remaining concave down until the inflection point at \( (1, 1) \). After this point, the function continues to decrease until reaching the local minimum at \( (2, 0) \), and then it begins to increase while being concave up.

In summary, to sketch a function's graph effectively, begin with the basics: determine the domain, intercepts, and symmetry. Then, utilize the first and second derivatives to analyze increasing/decreasing behavior and concavity. Finally, connect the plotted points smoothly to create an accurate representation of the function.

Summary of Curve Sketching Example 1

To sketch the graph of the rational function \( f(x) = \frac{x - 1}{x + 2} \), we will follow a systematic approach that includes determining the domain, intercepts, asymptotes, and analyzing the function's behavior using derivatives.

First, we establish the domain of the function. Since the denominator cannot be zero, we set \( x + 2 \neq 0 \), leading to the restriction \( x \neq -2 \). Therefore, the domain is \( (-\infty, -2) \cup (-2, \infty) \).

Next, we find the x-intercept by setting the numerator equal to zero: \( x - 1 = 0 \) gives \( x = 1 \). The y-intercept is found by evaluating \( f(0) = \frac{0 - 1}{0 + 2} = -\frac{1}{2} \). Thus, the intercepts are at \( (1, 0) \) and \( (0, -\frac{1}{2}) \).

Moving on to asymptotes, we identify the vertical asymptote by setting the denominator to zero: \( x + 2 = 0 \) results in \( x = -2 \). For the horizontal asymptote, we compare the degrees of the numerator and denominator. Both are of degree 1, so the horizontal asymptote is given by the ratio of the leading coefficients, which is \( y = 1 \).

Next, we check for symmetry by substituting \( -x \) into the function. The resulting expression does not match the original function or its negative, indicating that the function is neither even nor odd.

To analyze the function's increasing or decreasing behavior, we compute the first derivative using the quotient rule. The first derivative is given by:

\[ f'(x) = \frac{(x + 2)(1) - (x - 1)(1)}{(x + 2)^2} = \frac{3}{(x + 2)^2} \]

Since the numerator is always positive and the denominator is positive for \( x \neq -2 \), the function is increasing on its entire domain.

Next, we find the concavity by calculating the second derivative:

\[ f''(x) = \frac{d}{dx}\left(\frac{3}{(x + 2)^2}\right) = -\frac{6}{(x + 2)^3} \]

The second derivative is negative for \( x > -2 \) and positive for \( x < -2 \), indicating that the function is concave up on \( (-\infty, -2) \) and concave down on \( (-2, \infty) \). However, since \( x = -2 \) is an asymptote, it does not represent an inflection point.

Finally, we summarize the behavior of the function. It is increasing throughout its domain, with a vertical asymptote at \( x = -2 \) and a horizontal asymptote at \( y = 1 \). The function is concave up before the asymptote and concave down after it. To better visualize the graph, we can plot additional points, such as \( f(-3) = 4 \) and \( f(-4) = 2.5 \), to confirm the curve's behavior near the asymptotes.

In conclusion, the graph of \( f(x) = \frac{x - 1}{x + 2} \) will show an increasing function that approaches the vertical asymptote at \( x = -2 \) and the horizontal asymptote at \( y = 1 \), with distinct concavity changes around the asymptote.

Summary of Curve Sketching Example 2

To graph the function \( g(x) = \sqrt{x^3 + 8} \), we begin by determining its domain. Since the expression is under a square root, it must be non-negative, leading to the inequality \( x^3 + 8 \geq 0 \). Solving this gives \( x \geq -2 \), indicating that the domain of the function is \( [-2, \infty) \).

Next, we find the x- and y-intercepts. For the x-intercept, we set the function equal to zero: \( \sqrt{x^3 + 8} = 0 \). This simplifies to \( x^3 + 8 = 0 \), resulting in \( x = -2 \). For the y-intercept, we evaluate \( g(0) = \sqrt{0^3 + 8} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \). Thus, the intercepts are at \( (-2, 0) \) and \( (0, 2\sqrt{2}) \).

As we analyze the function for asymptotes, we find that there are none, as square root functions do not exhibit asymptotic behavior. To check for symmetry, we substitute \( -x \) into the function: \( g(-x) = \sqrt{(-x)^3 + 8} = \sqrt{-x^3 + 8} \). This does not equal \( g(x) \) or \( -g(x) \), indicating that the function is neither even nor odd.

Next, we explore the behavior of the function by finding its first derivative. Rewriting the function as \( g(x) = (x^3 + 8)^{1/2} \), we apply the power rule and chain rule to find:

\( g'(x) = \frac{1}{2}(x^3 + 8)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 8}}. \)

To find critical points, we set the numerator equal to zero: \( 3x^2 = 0 \), yielding \( x = 0 \). The derivative does not exist at \( x = -2 \), but since this point is not in the domain, we only consider \( x = 0 \). The intervals to test for increasing or decreasing behavior are \( [-2, 0) \) and \( (0, \infty) \). Testing values in these intervals shows that \( g'(x) > 0 \) for both intervals, indicating that the function is increasing throughout its domain.

To analyze concavity, we compute the second derivative using the quotient rule. The second derivative simplifies to:

\( g''(x) = \frac{3x(x^3 + 32)}{4(x^3 + 8)\sqrt{x^3 + 8}}. \)

Setting the numerator equal to zero gives potential inflection points at \( x = 0 \) and \( x^3 + 32 = 0 \) (the latter being outside the domain). The second derivative does not exist at \( x = -2 \). Testing intervals around these points reveals that the function is concave down on \( [-2, 0) \) and concave up on \( (0, \infty) \).

In summary, the function is increasing throughout its domain, with a point of inflection at \( x = 0 \). The graph starts at \( (-2, 0) \), increases while concave down until \( x = 0 \), and then continues to increase while concave up. The final sketch reflects these characteristics, showing a smooth curve that transitions from concave down to concave up at the inflection point.

Summary of Curve Sketching Example 3

To sketch a function based on given properties, we need to consider its continuity, differentiability, x-intercepts, and the behavior of its first and second derivatives. The function must be continuous and differentiable everywhere, which means it has no breaks or sharp corners.

We start by identifying the x-intercepts, which are points where the function crosses the x-axis. In this case, the function has x-intercepts at \( x = 0 \) and \( x = 2 \). These points can be plotted directly on the graph.

Next, we analyze the first derivative, which indicates where the function is increasing or decreasing. The first derivative is positive (greater than 0) from \( -\infty \) to \( 1 \) and from \( 2 \) to \( \infty \), indicating that the function is increasing in these intervals. Conversely, the function is decreasing from \( 1 \) to \( 2 \) where the first derivative is negative (less than 0).

Now, we turn to the second derivative, which tells us about the concavity of the function. The second derivative is positive (greater than 0) when \( x < 0.5 \) and \( x > 1.5 \), indicating that the function is concave up in these intervals. It is concave down (less than 0) from \( 0.5 \) to \( 1.5 \).

With this information, we can mark points of inflection where the concavity changes. The function changes from concave up to concave down at \( x = 0.5 \) and from concave down to concave up at \( x = 1.5 \). These points can be marked on the graph, and we can also identify local maxima and minima. The transition from increasing to decreasing at \( x = 1 \) indicates a local maximum, while the transition from decreasing to increasing at \( x = 2 \) indicates a local minimum.

Now, we can begin sketching the graph. Starting from the left, the function is increasing and concave up until \( x = 0.5 \). After this point, it continues to increase but becomes concave down until it reaches the local maximum at \( x = 1 \). The function then decreases until \( x = 2 \), where it reaches the local minimum (which is also an x-intercept). Finally, the function increases again and is concave up from \( x = 1.5 \) onward.

Throughout the sketch, we ensure that the function remains continuous and differentiable, adhering to the properties derived from the first and second derivatives. The final graph should reflect all these characteristics, confirming that the function is continuous, has the specified x-intercepts, and follows the increasing/decreasing and concavity behaviors as outlined.

Here’s what students ask on this topic:

How do you find the critical points of a function using the first derivative?

To find the critical points of a function, you need to take the first derivative of the function and set it equal to zero. Solve for the values of x where the first derivative is zero or does not exist. These x-values are the critical points. For example, for the function f(x) = x³ - 3x² + 4, the first derivative is f'(x) = 3x² - 6x. Setting this equal to zero, we get 3x(x - 2) = 0, which gives us critical points at x = 0 and x = 2.

What is the significance of the second derivative in curve sketching?

The second derivative of a function provides information about the concavity of the graph. If the second derivative is positive, the function is concave up (shaped like a smile). If the second derivative is negative, the function is concave down (shaped like a frown). Additionally, points where the second derivative changes sign are called inflection points, indicating a change in concavity. For example, for the function f(x) = x³ - 3x² + 4, the second derivative is f''(x) = 6x - 6. Setting this equal to zero, we find an inflection point at x = 1.

How do you determine the intervals where a function is increasing or decreasing?

To determine the intervals where a function is increasing or decreasing, you need to analyze the sign of the first derivative. First, find the critical points by setting the first derivative equal to zero. Then, create a sign chart by selecting test points in the intervals between the critical points. Plug these test points into the first derivative to determine its sign. If the first derivative is positive, the function is increasing in that interval. If it is negative, the function is decreasing. For example, for f(x) = x³ - 3x² + 4, the first derivative is f'(x) = 3x² - 6x, with critical points at x = 0 and x = 2. The function is increasing on (-∞, 0) and (2, ∞), and decreasing on (0, 2).

What are inflection points and how do you find them?

Inflection points are points on the graph of a function where the concavity changes. To find inflection points, take the second derivative of the function and set it equal to zero. Solve for the x-values where the second derivative is zero or does not exist. These x-values are potential inflection points. Verify by checking if the concavity changes on either side of these points. For example, for f(x) = x³ - 3x² + 4, the second derivative is f''(x) = 6x - 6. Setting this equal to zero, we get x = 1. Checking the concavity on either side of x = 1 confirms it is an inflection point.

How do you use the first and second derivative tests to find local extrema?

To find local extrema using the first derivative test, identify the critical points by setting the first derivative equal to zero. Analyze the sign of the first derivative around these points. If the first derivative changes from positive to negative, the function has a local maximum. If it changes from negative to positive, the function has a local minimum. The second derivative test involves evaluating the second derivative at the critical points. If the second derivative is positive, the function has a local minimum. If it is negative, the function has a local maximum. For example, for f(x) = x³ - 3x² + 4, the first derivative is f'(x) = 3x² - 6x, with critical points at x = 0 and x = 2. The second derivative is f''(x) = 6x - 6. Evaluating at x = 0 and x = 2 confirms the local extrema.