Problem 36 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=x \ln x-x $$

Step-by-Step Solution

Answer

The domain of the function \(f(x) = x \ln x - x\) is \((0,+\infty)\), and the range is \((-\infty,+\infty)\). The graph has an x-intercept at \(x=1\) and is decreasing on the interval \((0,1)\). It has a local minimum at \(x=1\) and is increasing on the interval \((1,+\infty)\). The graph is concave up for all values of x, with no inflection points or horizontal asymptotes.

Step 1: Domain and Range

The function is given by \(f(x)=x \ln x-x\). The natural logarithm has a domain of \(x>0\), and since it is multiplied by \(x\), the domain of the function will be the same. Thus, the domain is \((0,+\infty)\). The range of the function will be all real numbers, i.e., \((-\infty,+\infty)\).

Step 2: Intercepts

To find the x-intercepts, we set \(f(x)=0\): \(x \ln x - x = 0\) We factor out an \(x\): \(x(\ln x - 1) = 0\) This gives us the x-intercept: \(x=1\) The y-intercept can't be found, since the domain states that \(x>0\).

Step 3: Symmetry

Since the function is neither even nor odd, there is no symmetry.

Step 4: Increasing/Decreasing Intervals and Extrema

To find the increasing and decreasing intervals of the function, we consider the first derivative: \(f'(x)=\frac{d}{dx}(x \ln x-x)=\ln x+1-1=\ln x\) Now, we find where the derivative is zero or undefined: \(\ln x = 0 \Rightarrow x=e^0=1\) There is only one critical point: \(x=1\). Since \(\ln x\) is increasing, we have \(f'(x)<0\) for \(00\) for \(x>1\). Hence, the function is decreasing on the interval \((0,1)\) and increasing on the interval \((1,+\infty)\), with a local minimum at \(x=1\).

Step 5: Asymptotes

The function does not have vertical asymptotes, as the natural logarithm has a domain of \(x>0\). There are also no horizontal asymptotes.

Step 6: Concavity and Inflection Points

To determine concavity, we take the second derivative of the function: \(f''(x)=\frac{d^2}{dx^2}(x \ln x-x)=\frac{d}{dx}(\ln x)=\frac{1}{x}\) Now, we find where the second derivative is equal to zero or undefined and observe the changes in sign: - \(x<1: f''(x)>0\), which means the graph is concave up. - \(x>1: f''(x)>0\), which means the graph is still concave up. There are no inflection points, as the concavity does not change. Using the information gathered in steps 1-6, we can now sketch the graph of the function \(f(x)=x \ln x-x\). The function will have a domain of \((0,+\infty)\), a range of \((-\infty,+\infty)\), an x-intercept at \(x=1\), be decreasing on the interval \((0,1)\) and increasing on the interval \((1,+\infty)\) with a local minimum at \(x=1\), and will not change concavity.

Key Concepts

Domain and Range of Logarithmic Functions

Understanding the domain and range of logarithmic functions is crucial when sketching their graphs. In the case of the function \(f(x)=x \ln x - x\), we observe that it involves the natural logarithm \(\ln x\), which is defined only for positive values of \(x\). Therefore, the domain of \(f(x)\) is \((0, +\infty)\), meaning the function only accepts positive real numbers as input.

As for the range, since logarithmic functions can take on any real number as their output, and the linear term \(-x\) ensures that the function can decrease without limit, the range of \(f(x)\) is \((-\textbackslash note{∞}, +\textbackslash note{∞})\), allowing for any real number as the output. This understanding shapes our graph and is fundamental to plotting other characteristics of the function such as intercepts or asymptotes.

First Derivative Test for Increasing and Decreasing

The first derivative test is essential for determining where a function is increasing or decreasing. For the function in question, we calculated the first derivative as \(f'(x)= \ln x\). To apply the test, we need to examine where \(f'(x)\) is positive, negative, or zero, which indicates the growth or decline of the original function.

By setting \(\ln x = 0\), we found the critical point \(x = 1\). Observing the sign of \(f'(x)\), the function \(f(x)\) is decreasing when \(01\). At \(x=1\), there is a transition from decreasing to increasing, indicating a potential local minimum. Visualizing these intervals on a graph helps pinpoint where the function rises and falls.

Concavity and Inflection Points

Analyzing the concavity and locating possible inflection points in the graph of a function provide insight into its curvature. By considering the second derivative \(f''(x)=\frac{1}{x}\), we note that it never becomes zero nor changes sign, implying that the function has a consistent concavity.

In this specific function, \(f''(x)>0\) for all \(x>0\), which means the graph is always concave up. An inflection point is where the concavity changes from up to down or vice versa; however, given that there is no such change for this function, we conclude there are no inflection points. This information is particularly valuable for sketching the overall shape of the function on a graph.