Curve Sketching: Videos & Practice Problems

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 yields the x-intercepts at \( x = 2 \) and \( x = -1 \). The y-intercept is found by evaluating \( f(0) \), resulting in \( f(0) = 4 \), 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 is not equivalent to 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 \( x = 1 \). We analyze the intervals \( (-\infty, 1) \) and \( (1, \infty) \) using test values. The second derivative is negative on \( (-\infty, 1) \) (indicating concave down) and positive on \( (1, \infty) \) (indicating concave up). Thus, \( x = 1 \) is an inflection point.

To find local extrema, we apply the first derivative test. 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: \( (0, 4) \) is a local maximum and \( (2, 0) \) is a local minimum.

With all this information, we can 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, 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 for the rest of the graph.

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

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: \[f'(x) = \frac{(x + 2)(1) - (x - 1)(1)}{(x + 2)^2} = \frac{3}{(x + 2)^2}.\]Since the numerator is always positive and the denominator is positive for \( x \neq -2 \), \( f'(x) > 0 \) everywhere in its domain, indicating that the function is increasing on \( (-\infty, -2) \) and \( (-2, \infty) \).

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 \( (-2, \infty) \). The point \( x = -2 \) is not an inflection point since it is an asymptote.

Finally, we summarize the behavior of the function: it is increasing throughout its domain, concave up until \( x = -2 \), and concave down thereafter. We can plot additional points, such as \( f(-3) = 4 \) and \( f(-4) = 2.5 \), to better understand the curve's shape. The graph approaches the vertical asymptote at \( x = -2 \) and the horizontal asymptote at \( y = 1 \). The overall sketch will show the function increasing from the left, curving up to the asymptote, and then continuing to increase while curving downwards as it approaches the horizontal asymptote.

In conclusion, the graph of \( f(x) = \frac{x - 1}{x + 2} \) features a vertical asymptote at \( x = -2 \), a horizontal asymptote at \( y = 1 \), and is characterized by increasing behavior across its domain with a change in concavity at the asymptote.

To graph the function \( g(x) = \sqrt{x^3 + 8} \), we begin by determining its domain. Since the expression under the square root must be non-negative, we set up the inequality \( x^3 + 8 \geq 0 \). Solving this gives us \( x \geq -2 \), 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 leads 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 \( (0, 2\sqrt{2}) \).

Since \( g(x) \) is a square root function, it does not have any asymptotes. To analyze symmetry, we check \( g(-x) \). We find that \( g(-x) = \sqrt{-x^3 + 8} \), which is not equal to \( g(x) \) or \( -g(x) \), indicating that the function is neither even nor odd.

Next, we examine the function's behavior by finding its first derivative. Rewriting the function as \( g(x) = (x^3 + 8)^{1/2} \), we apply the 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, yielding \( 3x^2 = 0 \) or \( x = 0 \). The derivative does not exist at \( x = -2 \), but since this point is not in the domain, we only consider \( x = 0 \). We analyze the sign of \( g'(x) \) in the intervals \( [-2, 0) \) and \( (0, \infty) \). Testing values, we find that \( g'(x) > 0 \) in both intervals, indicating that the function is increasing throughout its domain.

Next, we find the second derivative to determine concavity. Using the quotient rule on \( g'(x) \), we derive:

\( g''(x) = \frac{(2\sqrt{x^3 + 8})(6x) - (3x^2)(3x^2)}{(2\sqrt{x^3 + 8})^2}. \)

After simplification, we find:

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

Setting the numerator equal to zero gives potential inflection points at \( x = 0 \) and \( x^3 + 32 = 0 \) (the latter is not in the domain). The second derivative does not exist at \( x = -2 \). Testing intervals around these points, we find that \( g''(x) < 0 \) for \( x \in [-2, 0) \) (concave down) and \( g''(x) > 0 \) for \( x \in (0, \infty) \) (concave up).

Since the first derivative does not change sign, there are no local extrema. Finally, we sketch the graph, noting that it starts at \( (-2, 0) \), is concave down and increasing until \( x = 0 \), and then becomes concave up while continuing to increase. The resulting graph reflects these characteristics, showing a smooth curve that transitions from concave down to concave up at \( x = 0 \).

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).

Moving on to the second derivative, which tells us about the concavity of the function, we find that it is positive (concave up) when \( x < 0.5 \) and \( x > 1.5 \). This means the function is concave up in these intervals, resembling a "smile." In contrast, the function is concave down (negative second derivative) between \( 0.5 \) and \( 1.5 \), resembling a "frown."

With this information, we can mark points of inflection where the concavity changes. The first inflection point occurs at \( x = 0.5 \) (from concave up to down), and the second at \( x = 1.5 \) (from concave down to up). These points can be placed at any y-value, as long as the final graph adheres to the properties outlined.

Next, we 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. The local maximum will be above the other points, while the local minimum coincides with the x-intercept at \( x = 2 \).

Now, we can begin sketching the graph. Starting from the left, the function is concave up and increasing until it reaches the inflection point at \( x = 0.5 \). After this point, it becomes concave down and continues to increase until it reaches the local maximum at \( x = 1 \). The function then decreases until it reaches the local minimum at \( x = 2 \), where it changes concavity again to concave up and increases towards infinity.

Throughout this process, we ensure that the function remains continuous and differentiable, with the correct behavior of the first and second derivatives. The final graph should reflect all these properties, confirming that the function behaves as expected based on the given criteria.