Question

a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval. b. Use a graphing utility to find all the solutions to the equation on the given interval. c. Illustrate your answers with an appropriate graph. $$x\ln x-1=0;(1,e)$$

Short answer

Question: Show that the equation $x\ln x-1=0$ has a solution on the interval $(1,e)$. Answer: The equation $x\ln x-1=0$ has a solution on the interval $(1,e)$ because the function $f(x)=x\ln x-1$ is continuous on this interval, and by the Intermediate Value Theorem, there must exist a value $k$ in $(1,e)$ such that $f(k)=0$. A graph of the function illustrates this by showing the function crossing the $x$-axis at an approximate value of $x\approx 2.256$ in the specified interval.

Step-by-step solution

Check continuity on the interval (1, e)

The given function is $f(x)=x\ln x-1$. Since both $\ln x$ and $x$ are continuous functions, their product is also continuous and thus $f(x)$ is continuous on the interval $(1,e)$.

Find values at the endpoints of the interval (1, e)

Now we will find the values of the function at the endpoints of the interval $(1,e)$. At $x=1$: $$f(1)=1\ln 1-1=0-1=-1$$ At $x=e$: $$f(e)=e\ln e-1=e\times 1-1=e-1$$ Since $e>1$, we have $e-1>0$. Therefore, $f(1)=-1<0$ and $f(e)=e-1>0$.

Apply the Intermediate Value Theorem

To apply the Intermediate Value Theorem, we need to show that the function is continuous on the interval $[1,e]$ and that there is a value $k$ in $(1,e)$ for which $f(k)=0$. Since the function is continuous, and $f(1)<0$ and $f(e)>0$, by the Intermediate Value Theorem, there must exist a value $k$ in $(1,e)$ such that $f(k)=0$. This is because the function will pass through zero as it moves from a negative value to a positive value without any discontinuity. Thus, the equation $x\ln x-1=0$ has a solution on the interval $(1,e)$.

Use a graphing utility to find solutions

Now we will use a graphing utility to find all the solutions of the equation on the interval $(1,e)$. When graphing the function $f(x)=x\ln x-1$ on the interval $(1,e)$, we find that there is a root at $x\approx 2.256$, which indicates that the equation $x\ln x-1=0$ has a solution in the interval $(1,e)$.

Illustrate with a graph

Plot the function $f(x)=x\ln x-1$ on the interval $(1,e)$. The graph will show the behavior of the function in the given interval and how it crosses the $x$-axis at the approximate value $x\approx 2.256$. This visualizes that there indeed is a solution to the equation $x\ln x-1=0$ on the interval $(1,e)$.

Key concepts

Continuity of Functions

Continuity of a function is a fundamental concept in calculus that describes a smooth, unbroken path of values the function takes as the input changes. If a function is continuous at a point, it means that there's no abrupt change or gap in its graph at that point, and you can draw its graph without lifting your pencil.

More formally, a function is continuous over an interval if it’s continuous at every point within that interval. This involves three conditions: the function is defined at the point, the limit exists at that point, and the function’s value equals the limit. When looking at our exercise involving the Intermediate Value Theorem (IVT), assessing continuity becomes crucial. In the provided example, the function f(x) = x \text{log}x - 1 is continuous over the interval (1, e) because both $\text{log}x$ and $x$ are continuous, and the operations of multiplication and subtraction (which compose our function) do not disrupt this continuity.

Why is this so essential? Well, the IVT hinges on continuity. It states that for any function continuous on [a, b], and any value between f(a) and f(b), there exists a point c in [a, b] where f(c) equals that value. Intuitively, it means that if you know a function starts below the x-axis (negative value) and ends above it (positive value) without any breaks, it must have crossed the axis somewhere in between. So, by showing that our function is continuous over (1, e), we’ve laid the groundwork for employing the IVT and guaranteeing a solution exists within that interval.

Graphing Utilities

Graphing utilities, such as graphing calculators or computer algebra systems, are indispensable tools in exploring the behavior of functions and finding solutions to equations. They allow students to visualize the form and key features of functions – including intercepts, peaks, troughs, and points of inflection – quickly and accurately.

When we use these graphing utilities in the context of the given problem, our main goal is to pinpoint where the function f(x) = x \text{log}x - 1 intersects the x-axis within the interval (1, e). This interception point is where the function’s value is zero, meaning it represents the solution to the equation $x\text{log}x-1=0$. With these digital tools, we can zoom in on the interval of interest and observe that the function crosses the x-axis around $x\approx 2.256$, thus finding our approximate solution.

When writing about graphing utilities, it is helpful for students to understand that these are not just abstract tools but practical aids that complement analytical methods, providing both visual insight and supplementary data for understanding and solving mathematical problems.

Solving Equations

The process of solving equations is at the core of algebra and, by extension, many other areas of mathematics. It involves finding the values that make an equation true – in other words, the points where one side of the equation equals the other. There are numerous methods to solve equations, ranging from algebraic manipulation and factorization to graphical and numerical approaches.

In our case, we are seeking to solve the transcendental equation $x\text{log}x-1=0$. Solving an equation like this analytically can often be quite challenging due to the involvement of both a logarithm and algebraic terms. Therefore, we can apply a mix of numerical methods and graphing to find solutions. The use of graphing utilities is an effective approach to numerically solving equations as it provides visual confirmation for where the solutions might lie.

Once a function is graphed, we look for the x-values at the points where the graph intersects the x-axis, since these points correspond to the solutions of the equation. For the equation within our exercise, the graph reveals a solution near $x\approx 2.256$. This not only provides an approximate answer but also reinforces a deeper understanding of how a function behaves, leading to a solution by observing where the function reaches the value of zero.