/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Find the function The following ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 57

Find the function The following limits represent the slope of a curve \(y=f(x)\) at the point \((a, f(a)) .\) Determine a possible function \(f\) and number a; then calculate the limit. $$\lim _{x \rightarrow 2} \frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}$$

Short Answer

Expert verified
#Answer# The function \(f(x)\) is given by: $$f(x) = \ln |x+1| - \frac{1}{3}x + C$$ and the value of the limit is: $$\lim _{x \rightarrow 2} \frac{\frac{1}{x+1}-\frac{1}{3}}{x-2} = 0$$

Step by step solution

01

Write the limit expression as a derivative

The given limit expression can be written as the derivative of \(f\) at \(x=a=2\): $$f'(a) = \lim _{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$$ where \(f(x) = \frac{1}{x+1}\) and \(a=2\). Now, we have: $$f'(2) = \lim _{x \rightarrow 2} \frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}$$
02

Integrate the function f'(x)

To find the function \(f(x)\), we can integrate its derivative, \(f'(x)\): $$f(x) = \int f'(x) dx$$ Notice that since the given limit represents the slope of a curve at the point \((2, f(2))\), we are looking for a function whose derivative is of the form: $$f'(x) = \frac{1}{x+1} - \frac{1}{3}$$ Integrating this expression gives us: $$f(x) = \int \Big(\frac{1}{x+1} - \frac{1}{3}\Big) dx$$ $$f(x) = \int \frac{1}{x+1} dx - \int \frac{1}{3} dx$$ $$f(x) = \ln |x+1| - \frac{1}{3}x + C$$
03

Calculate the limit

Now that we have found the function \(f(x)\), we can determine the given limit: $$\lim _{x \rightarrow 2} \frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}$$ As \(x\) approaches \(2\), the function \(\frac{1}{x+1}\) approaches \(\frac{1}{2+1} = \frac{1}{3}\). Therefore: $$\lim _{x \rightarrow 2} \frac{\frac{1}{3}-\frac{1}{3}}{x-2} = \lim _{x \rightarrow 2} \frac{0}{x-2} = 0$$ So, the given limit equals zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics that studies how things change. It helps us to understand the behavior of functions and provides tools to analyze complex problems involving motion, area, volume, and other dynamic concepts. At its core, calculus is divided into two main areas: differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which describes the rate at which things change, while integral calculus is about finding the total accumulation of quantities. In our exercise, we see these principles in action as we explore the rate of change of a function at a specific point and how we can reverse this process to find the original function.
Derivative as Slope
In calculus, the derivative represents the slope of a function at a given point. The slope is a measure of how steep a line is, which in calculus translates to how quickly a function is changing at that point. The concept is similar to the slope you learned in algebra, but instead of a straight line, the derivative deals with slopes of curves. For the function f(x), the derivative f'(x) at x=a is the limit of the difference quotient as x approaches a.

The given limit in our exercise can be seen as the derivative f'(a) of the function f(x) at the point a=2. Thus, it represents the slope of the tangent to the curve at that specific point, showing how the function changes as x moves away from 2.
Integrating Derivatives
Integrating a derivative essentially means finding the original function from its rate of change. This process is known as anti-differentiation or integration. When we integrate a derivative, we are adding up all the infinitesimal changes (represented by the derivative) to get the total change, which gives us the function itself. In our example, by integrating the derivative f'(x), we derive the original function f(x). The integral symbol 鈭� represents this accumulation of change, and since the derivative is the opposite of integration, we can recover the original function by performing an integral on the derivative.
Limits of Functions
The concept of limits is fundamental in calculus and is used to describe the behavior of functions as they approach a specific point. A limit can tell us the value that a function approaches as the input (or x-value) gets infinitely close to some number. Limits are essential in defining the derivative and the integral. They allow us to handle situations where functions might otherwise be undefined, such as division by zero.

In our exercise, we calculate the limit as x approaches 2 for the slope of the function f(x). This limit represents the exact slope of the tangent line to the curve at the point where x equals 2. As it turns out, this particular limit is zero, indicating that at x=2, the function doesn't change; the tangent line to the curve at that point is flat.

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Most popular questions from this chapter

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(2 x)^{2 x}$$.

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general, $$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\)

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$
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