Question

Use a graphing utility together with analytical methods to create a complete graph of the following functions. Be sure to find and label the intercepts, local extrema, inflection points, asymptotes, intervals where the function is increasing/decreasing, and intervals of concavity. $$f(x)=x / \ln x$$

Short answer

Based on the analysis of the function $$f(x) = \frac{x}{\ln x}$$, determine if the following statement is true or false: The function has a local minimum at the point $$(e,e)$$. Answer: True

Step-by-step solution

Analyze the function and its domain

First, let's determine the domain of the function $$f(x)=\frac{x}{\ln x}$$. Since the denominator is the natural logarithm, we cannot have $$\ln x = 0$$, which means $$x \neq 1$$. Additionally, the natural logarithm is undefined for $$x \leq 0$$. Therefore, the domain of the function is $$x \in (0, 1) \cup (1, +\infty)$$. Now, we will find the main features of the function.

Find the intercepts

To find the x-intercept, we set $$f(x) = 0$$: $$0 = \frac{x}{\ln x}$$. However, since the function is undefined for $$x=1$$, we cannot find an x-intercept. To find the y-intercept, we set $$x = 0$$. However, as noted earlier, the function is also undefined at $$x = 0$$. Therefore, there is no y-intercept either.

Find the local extrema

To find the local extrema, we need to find the critical points by setting the first derivative equal to zero. The first derivative of f(x) is: $$f'(x) = \frac{\ln x - 1}{(\ln x)^2}$$. Set $$f'(x)=0$$, we get $$\ln x = 1, x = e$$. This is our only critical point. To determine if it's a maximum or minimum, we can use the second derivative test: $$f''(x) = \frac{2 - \ln x}{(\ln x)^3}$$. Substitute the critical point $$x = e$$ into the second derivative: $$f''(e) = \frac{2 - 1}{1^3} = 1$$, which is positive. Therefore, there is a local minimum at $$(e, \frac{e}{\ln {e}}) = (e,e)$$.

Find the inflection points

To find the inflection points, we need to set the second derivative to zero: $$0 = \frac{2 - \ln x}{(\ln x)^3}$$. This gives us $$\ln x = 2, x = e^2$$. Thus, there is an inflection point at $$(e^2, \frac{e^2}{\ln {e^2}}) = (e^2, 2e^2)$$.

Asymptotes and increasing/decreasing intervals

Since the function is a rational function with the denominator higher than the numerator, there is a horizontal asymptote at $$y=0$$. Additionally, there is a vertical asymptote at $$x=1$$, where the function is undefined. The function is increasing wherever the first derivative is positive and decreasing where the derivative is negative: - Increasing: For $$0 < x < e$$ and $$x>e^2$$ - Decreasing: For $$e < x < e^2$$

Determine the intervals of concavity

The function is concave up wherever the second derivative is positive and concave down where the derivative is negative: - Concave up: For $$x < e^2$$ - Concave down: For $$x > e^2$$ We have now found and labeled all the required features for the given function. To further visualize the function, it's a good idea to use a graphing utility to plot the function and check our work.

Key concepts

Graphing Functions

When graphing a function like \(f(x) = \frac{x}{\ln x}\), it's crucial to understand its shape and key characteristics. Graphing utilities can help visualize these functions and confirm analytical findings. By plotting the function, you can see where it moves upwards, downwards, and where it changes direction. Before you start graphing, it's essential to find the domain: the set of x-values where the function is defined. For \(f(x) = \frac{x}{\ln x}\), the domain is \(x \in (0, 1) \cup (1, +\infty)\). Knowing the domain helps you avoid graphing at undefined points. After determining the domain, look for significant aspects like intercepts, extrema, and intervals of increase or decrease. Graphing functions is a blend of art and mathematics, combining calculated predictions with actual visual graphs.

Local Extrema

Local extrema refer to the points where the function reaches a local maximum or minimum. These are the 'peaks' or 'valleys' within a given interval. For the function \(f(x) = \frac{x}{\ln x}\), local extrema are found by taking the first derivative and solving for zeros or undefined points. After finding these critical points, like \(x = e\), you use tests, such as the second derivative test, to determine whether these are maxima or minima. In our example, at \(x = e\), the value of the second derivative is positive, indicating a local minimum. These points are important as they reveal where the function changes direction from increasing to decreasing, or vice versa.

Inflection Points

An inflection point is where the curve changes concavity, or in simpler terms, switches from being 'bowl-shaped' to 'hat-shaped'. For finding these in \(f(x) = \frac{x}{\ln x}\), we look at the second derivative. Setting \(f''(x) = 0\), we can find points where the concavity changes. For instance, solving yields \(x = e^2\), indicating an inflection point at \((e^2, 2e^2)\). Identifying inflection points helps us understand the broader shape of the graph and areas where it shifts from concave up to concave down. These changes help users predict and describe physical scenarios where this model might apply.

Asymptotes

Asymptotes are lines the graph approaches but never touches. In our example of \(f(x) = \frac{x}{\ln x}\), there are both horizontal and vertical asymptotes. The horizontal asymptote, where \(f(x)\) approaches as \(x\) tends to infinity, is \(y = 0\). This suggests the function never crosses the x-axis, but gets infinitely close to it. Simultaneously, the vertical asymptote at \(x = 1\) shows a point of discontinuity where the function is not defined. As \(x\) approaches this point from either side, the function value spikes towards infinity or drops sharply, creating a 'gap' in the graph. Understanding asymptotes is vital as they depict behavioral trends of functions at extreme domain values.

Intervals of Concavity

The intervals of concavity tell us whether a function is curving upwards or downwards. By examining the sign of the second derivative \(f''(x)\), we determine these intervals. For the function \(f(x) = \frac{x}{\ln x}\), it is concave up when \(f''(x) > 0\), specifically in the interval \(x < e^2\). Conversely, it is concave down when \(f''(x) < 0\) for \(x > e^2\). Identifying these areas helps in predicting the motion direction, making it easier to anticipate places where the function might shift unexpectedly. This understanding is beneficial in both theoretical calculus and practical applications, enabling better comprehension and prediction of natural phenomena.