Question

Graphing functions Make a complete graph of the following functions using the graphing guidelines outlined. $$f(x)=x \ln x$$

Short answer

Question: Sketch the graph of the function $$f(x) = x \ln x$$ and identify its key features, including domain, range, critical points, and asymptotes. Answer: The function $$f(x) = x \ln x$$ has domain $$(0, \infty)$$, range $$(-\infty, \infty)$$, a local minimum at $$x = e^{-1}$$, and a vertical asymptote at $$x = 0$$. The graph is concave up throughout its domain.

Step-by-step solution

Identify the Domain and Range

The function is defined for $$x \gt 0$$ since the natural logarithm is only defined for positive numbers. Therefore, the domain is $$(0, \infty)$$. The range of the function also includes all real numbers since the function will have both positive and negative values for different values of x in its domain.

Find the First Derivative

We will find the first derivative to determine the critical points (where the function changes its behavior from increasing to decreasing or vice versa) and check for any local maxima or minima. To do this, use the product rule: $$f'(x) = \frac{d}{dx}(x \ln x) = x \cdot \frac{1}{x}+\ln x = 1+\ln x$$

Find the Critical Points

To find the critical points, we need to find the values of x for which the first derivative is equal to zero or is undefined. In this case, the first derivative has no value for which it is undefined, so we only need to set it to zero to find the critical points: $$f'(x) = 0 \Rightarrow 1+\ln x = 0$$ Solving for x, we get one critical point: $$x = e^{-1}$$.

Determine the Behavior of the Function around the Critical Point

To see whether our critical point represents a local maximum or minimum, or a simple point of inflection, examine the behavior of the function at points on either side of the critical point using the first derivative: - Left of the critical point ($$x < e^{-1}$$): $$f'(x) = 1 + \ln x < 1+\ln e^{-1} = 0$$, so the function is decreasing. - Right of the critical point ($$x > e^{-1}$$): $$f'(x) = 1 + \ln x > 1+\ln e^{-1} = 0$$, so the function is increasing. Thus, we have a local minimum at $$x = e^{-1}$$.

Find the Second Derivative

Now find the second derivative to check for concavity and inflection points: $$f''(x) = \frac{d^2}{dx^2}(x \ln x) = \frac{d}{dx}(1+\ln x) = \frac{1}{x}$$

Determine the Concavity and Inflection Points

Since the second derivative is always positive when $$x > 0$$, the graph is concave up everywhere within its domain and there are no inflection points.

Vertical Asymptotes

As the function is not defined for $$x = 0$$ and $$\lim_{x\to0^+}x\ln x=-\infty$$, there is a vertical asymptote at $$x = 0$$.

Sketch the Graph

Taking into account all the information gathered in the previous steps, create a sketch of the graph of the function $$f(x) = x \ln x$$: - The graph has a vertical asymptote at $$x = 0$$. - The function has a local minimum at $$x = e^{-1}$$. - The graph is concave up throughout its domain. Sketch a curve with these characteristics, and make sure to label the axes with the appropriate information (domain, range, critical points, and asymptotes).

Key concepts

Domain and Range of a Function

Understanding the domain and range of a function is a fundamental concept in calculus and crucial when graphing functions. The domain refers to the set of all possible input values (x-values) for which the function is defined, while the range is the set of all possible output values (y-values) that the function can produce.

For the function f(x) = x \text{ln}(x), the logarithm requires that the input must be a positive real number. Therefore, the domain is \(0, \infty\). Considering the behavior of the function, as the value of x approaches zero from the right, the function's value approaches negative infinity, indicating that the range includes all real numbers. Understanding these concepts helps us anticipate the shape and position of a function's graph on the coordinate plane.

First Derivative Test

The first derivative test is an essential tool for analyzing the behavior of a function's graph around its critical points, where its slope changes. By taking the derivative of the function and setting it to zero, we can find these critical points. For the function \(f(x) = x \text{ln}(x)\), the first derivative is \(f'(x) = 1 + \text{ln}(x)\).

By solving \(f'(x) = 0\), we can establish the locations of potential maximums, minimums, or points of inflection. With \(f'(x)\), we've discovered that \(x = e^{-1}\) is a critical point. Investigating values just to the left and right of this point, we determine if this represents a local minimum or maximum, which informs us about the function's increasing or decreasing behavior in its vicinity.

Concavity and Inflection Points

The concavity of a function's graph relates to the direction it curves, and inflection points are where this curvature changes. Concavity is determined by the second derivative, \(f''(x)\). If \(f''(x) > 0\), the graph is concave up, and if \(f''(x) < 0\), it is concave down. Inflection points occur where \(f''(x)\) changes sign.

For our function \(f(x) = x \text{ln}(x)\), \(f''(x) = \frac{1}{x}\) is always positive for \(x > 0\). As a result, the graph is concave up in its entire domain, and there are no inflection points. This information gives us further insight into the graph's shape, showing us that it will always curve in the same manner—similar to the right-hand side of a parabola.

Vertical Asymptotes

When graphing a function, vertical asymptotes occur at values of x where the function approaches infinity (or negative infinity). These are values excluded from the domain of the function and represent boundaries beyond which the graph cannot cross.

In our specific example with \(f(x) = x \text{ln}(x)\), as x approaches 0 from the right, the function's value approaches negative infinity. Therefore, there is a vertical asymptote at \(x = 0\). Recognizing vertical asymptotes is vital because they are key indicators of the behavior of the graph near the edges of its domain.

Product Rule in Calculus

When differentiating functions, we often come across products of two or more functions. The product rule is a critical tool for managing such situations. It states that if you have a product of two functions, u(x) and v(x), the derivative of their product is \(u'(x) v(x) + u(x) v'(x)\).

In the function \(f(x) = x \text{ln}(x)\), using the product rule allows us to find the derivative efficiently. By letting \(u(x) = x\) and \(v(x) = \text{ln}(x)\), and applying the rule, we can find \(f'(x)\) and then use it for further analysis of the function's graph. The product rule is a fundamental derivative rule that has broad applications across calculus problems.