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Chapter 6: Problem 7

Nd a quadratic or linear function \(f(x)\) whose derivative is the line speci ed and whose graph passes through: (a) the origin, (b) the point \((0,2)\). (a) \(f^{\prime}(x)=6 x-2\) (b) \(f^{\prime}(x)=m x+b, m \neq 0\)

Short Answer

Expert verified
The original function for case (a) is \(f(x) = 3x^2 - 2x\) and for case (b) is \(f(x) = \frac{1}{2} m x^2 + bx + 2\).

Step by step solution

01

Finding the original function from the derivative - Case (a)

For given derivative function \(f^{\prime}(x)=6 x-2\), we need to find the integral of this function. By the power rule for integration, the integral of \(x\) is \(\frac{1}{n+1} x^{n + 1}\), where \(n\) is the power of \(x\). By adding the constant of integration \(C\), we get the original function. The integral of \(6 x - 2\) would be \(3 x^2 - 2x + C\).
02

Finding the constant of integration - Case (a)

Now that we have our original function as \(f(x) = 3x^2 - 2x + C\), we need to find \(C\). With the information that the function passes through the origin \((0,0)\), we plug these coordinates into our original function and solve for \(C\): \(0 = 3(0)^2 - 2(0) + C\), so \(C = 0\). So, our original function is \(f(x) = 3x^2 - 2x\).
03

Finding the original function for a general derivative - Case (b)

For given general derivative function \(f'(x) = m x + b\), where \(m\) is the coefficient of \(x\) and \(m\neq0\), we find the integral of this function. The integral of \(m x + b\) would be \(\frac{1}{2} m x^2 + bx + C' \), where \(C'\) is the constant of integration for case \(b\).
04

Finding the constant of integration - Case (b)

Now that we have our original function as \(f(x) = \frac{1}{2} m x^2 + bx + C'\), we need to find \(C'\). With the information that the function passes through point \((0,2)\), we plug these coordinates into our original function and solve for \(C'\): \(2 = \frac{1}{2} m (0)^2 + b(0) + C'\), so \(C' = 2\). So, our original function is \(f(x) = \frac{1}{2} m x^2 + bx + 2\).

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus, often thought of as the reverse process of differentiation. In simple terms, integration is used to find the original function from its derivative. When we integrate, we accumulate the quantity described by a given rate of change, finding a function that describes the entire accumulation.
Consider the integration of functions involving power of x. For example, when integrating a function like \(6x - 2\), we use the power rule of integration. The power rule states:
  • For any function \(x^n\), the integral is \(\frac{1}{n+1}x^{n+1} + C\),
  • where \(n\) is any real number, and \(C\) is the constant of integration.
In practice, this means applying the power rule to each term separately and adding the constant \(C\).
Hence, integrating \(6x - 2\), we get \(3x^2 - 2x + C\). This expression is our original function before the application of differentiation.
Quadratic Function
A quadratic function is a type of polynomial that can be written in the standard form \(ax^2 + bx + c\). When dealing with the quadratic function, the coefficient \(a\) determines the shape and direction of the parabola. If \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards.
Quadratic functions are essential in calculus as they frequently appear when determining areas and solving optimization problems. In the exercise described, we found the original quadratic function \(f(x) = 3x^2 - 2x\) from the derivative \(f'(x) = 6x - 2\). Here, the degree of the polynomial of the derivative informs us of the degree of the resulting function after integration.
In the exercise, we ensure the quadratic function meets certain conditions, namely passing through specific points like the origin and point \((0,2)\). This involvement of points helps in determining the constant in the general integral expression.
Derivative
The concept of the derivative is central in calculus, representing the instantaneous rate of change. Think of it as the slope of the tangent to a curve at any point. For any function \(f(x)\), the derivative \(f'(x)\) describes how \(f(x)\) changes as \(x\) changes.
When given a derivative like \(f'(x) = 6x - 2\), it indicates how the original function's slope changes for every increase in \(x\). This hints towards the behavior of the original function.
In practice, finding the actual function from its derivative involves integration, as shown in the integration process. This step is vital when you have information about the derivative and need to reconstruct the original function, including its specific behavior, such as passing through given points. By knowing the derivative, you already have a treasure map to the land of functions, guiding you on how steep or flat your journey will be.

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Most popular questions from this chapter

A seat on a round-trip charter flight to Cairo costs \(\$ 720\) plus a surcharge of \(\$ 10\) for every unsold seat on the airplane. (If there are 10 seats left unsold, the airline will charge each passenger \(\$ 720+\$ 100=\$ 820\) for the flight.) The plane seats 220 travelers and only round-trip tickets are sold on the charter flights. (a) Let \(x=\) the number of unsold seats on the flight. Express the revenue received for this charter flight as a function of the number of unsold seats. (Hint: Revenue = (price + surcharge)(number of people flying).) (b) Graph the revenue function. What, practically speaking, is the domain of the function? (c) Determine the number of unsold seats that will result in the maximum revenue for the flight. What is the maximum revenue for the flight?

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