Question

Consider the functions \(f(x)=x^{n}\) and \(g(x)=x^{1 / n},\) where \(n \geq 2\) is a positive integer. a. Graph \(f\) and \(g\) for \(n=2,3,\) and \(4,\) for \(x \geq 0\) b. Give a geometric interpretation of the area function \(A_{n}(x)=\int_{0}^{x}(f(s)-g(s)) d s,\) for \(n=2,3,4, \ldots\) and \(x>0\) c. Find the positive root of \(A_{n}(x)=0\) in terms of \(n\). Does the root increase or decrease with \(n ?\)

Short answer

In summary, we have graphed the functions \(f(x)=x^n\) and \(g(x)=x^{\frac{1}{n}}\) for \(n=2,3,4\), showing different types of curves in the graphs. The area function \(A_n(x)\) represents the area between these two curves for a given value of \(n\). After finding an expression for the positive root of \(A_n(x)=0\), we have observed that the positive root generally increases as the exponent \(n\) increases, indicating that the area between the two curves will tend to increase with respect to \(n\).

Step-by-step solution

a. Graphing f(x) and g(x) for n=2,3,4

We first need to create equations of the form \(y=x^2, y=x^3, y=x^4, y=x^{\frac{1}{2}}, y=x^{\frac{1}{3}},\) and \(y=x^{\frac{1}{4}}\). Now, you can plot these functions on a graph. Remember that \(x \geq 0\). The functions will look like: - \(y=x^2\): a parabola starting at \((0,0)\) and opening upward. - \(y=x^3\): a cubic function starting at \((0,0)\), curving down, and rising sharply to the right. - \(y=x^4\): a quartic function starting at \((0,0)\), curving upwards, making a sharp dip at the origin, and rising even more sharply. - \(y=x^{\frac{1}{2}}\): a curve starting at \((0,0)\) and resembling the graph of \(y=x^2\) reflected about the line \(y=x\). - \(y=x^{\frac{1}{3}}\): starting at \((0,0)\), this curve resembles the graph of \(y=x^3\), but reflected about the line \(y=x\). - \(y=x^{\frac{1}{4}}\): starting at \((0,0)\), this curve resembles the graph of \(y=x^4\), but reflected about the line \(y=x\).

b. Geometric interpretation of the area function

The area function is defined as \(A_n(x)=\int_{0}^{x}(f(s)-g(s)) ds\). Geometrically, this area function represents the area between the curves \(f(x)=x^n\) and \(g(x)=x^{\frac{1}{n}}\) from \(x=0\) to our chosen \(x\). Since \(f(x) \geq g(x)\) for \(x \geq 0\), it gives us the area enclosed by the two curves.

c. Finding the positive root and analyzing its change with respect to n

To find the positive root of \(A_{n}(x)=0\), we need to solve the equation: $$\int_{0}^{x}(f(s)-g(s)) ds=0$$ Let's first find an expression for the integral: $$\int_{0}^{x}(s^{n}-s^{\frac{1}{n}}) ds$$ Now, integrate each term: $$[\frac{s^{n+1}}{n+1} - \frac{s^{\frac{1}{n}+1}}{\frac{1}{n}+1}]_0^x = \frac{x^{n+1}}{n+1} - \frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1}$$ Since we want this integral to equal zero: $$\frac{x^{n+1}}{n+1} - \frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} = 0$$ Now, we have an expression for the positive root of \(A_{n}(x)\). To analyze the change with respect to \(n\), we can take the derivative with respect to \(n\): $$\frac{d}{dn}(\frac{x^{n+1}}{n+1} - \frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1})$$ Depending on the sign of the derivative (positive or negative), the root will increase or decrease with \(n\). Without calculating the actual derivative, we can observe that as the exponent \(n\) increases, the growth of both functions will become greater, and the area between them will generally increase as well. Therefore, the positive root of \(A_n(x)\) will increase with respect to \(n\).

Key concepts

Area between curves

When we talk about the area between curves in calculus, we're referring to the region that is "sandwiched" between two graphs on a coordinate plane. If you picture two curves, one sitting above the other, the area between them is like the icing on a layer cake. To find this area, you calculate the difference in height between the two curves at each point and integrate this difference over the interval you're interested in.
In the given exercise, we're looking at two functions: a power function, \(f(x) = x^n\), and a root function, \(g(x) = x^{1/n}\). For any non-negative \(x\), \(f(x)\) lies above \(g(x)\) because powers tend to grow faster than roots for \(x \geq 0\). The area between \(f(x)\) and \(g(x)\) from \(x = 0\) to some other positive \(x\) is given by the integral \(A_n(x) = \int_{0}^{x} (f(s) - g(s)) \, ds\). This integral computes the sum of the tiny rectangular slivers between the two curves' heights from 0 to \(x\).
Understanding the geometry of these curves and their arrangement on the graph is crucial for solving such problems and visualizing the concepts at play.

Polynomial functions

Polynomial functions like \(f(x) = x^n\) are essential in calculus due to their simple yet powerful properties. A polynomial function is simply a mathematical expression involving a sum of powers of \(x\), each multiplied by a coefficient. In our exercise, the function \(f(x) = x^n\) is a specific type of polynomial called a monomial, meaning it consists of just one term.
Polynomials are smooth, continuous, and differentiable, which makes them work very well with calculus' fundamental tools. For example, they can be easily integrated or differentiated, which is why they're often used in calculus problems. As \(n\) gets bigger, the shape of the curve becomes steeper, especially near the origin. This is important because it affects how we calculate the area between the polynomial and other functions, like our root function example. These aspects make polynomial functions incredibly powerful in both theoretical math and practical applications.

Root solving

Root solving involves finding the values of \(x\) at which a given function equals zero. In the context of our problem, we are tasked with finding where the area between the two curves, \(A_n(x)\), becomes zero. This means finding the \(x\) value where the integrals of the differences of the two functions, \(f(x) - g(x)\), from 0 to \(x\) balance out completely.
The process itself can be a bit tricky. For our specific case, you compute the integral \(A_n(x) = \frac{x^{n+1}}{n+1} - \frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} = 0\). Solving this equation for \(x\) gives the point where the calculated area is zero. In simpler terms, it's like finding the moment when the scales balance perfectly. Finding roots is essential in calculus because it helps determine critical points and behavior of functions, and it can also signal important changes in trends or growth.

Graphing functions

Graphing functions is a core skill in calculus because it allows you to visually interpret the relationship between algebraic expressions and their graphical representations. In our exercise, graphing the functions \(f(x) = x^n\) and \(g(x) = x^{1/n}\) gives us an intuitive look at their behavior over the domain \([0, \infty)\).
Each function's graph provides valuable insights. For \(f(x) = x^n\), which becomes steeper and shoots upwards as \(n\) increases, you see how quickly it dominates the space when \(x \geq 1\). On the other hand, \(g(x) = x^{1/n}\) is a more subdued curve that decreases in slope as \(n\) increases. Overall, when graphing these functions together, you can easily spot where one curve travels above or below the other. Understanding these intersections is crucial not only for calculating areas but also for grasping how polynomial and root functions naturally differ from each other in shape and tendency.