/*! 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 37 Let \(g(x)=x^{3}-4 x+6 .\) Find ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 37

Let \(g(x)=x^{3}-4 x+6 .\) Find \(f(x)\) so that \(f^{\prime}(x)=g^{\prime}(x)\) and \(f(1)=2\).

Short Answer

Expert verified
\( f(x) = x^3 - 4x + 5 \)

Step by step solution

01

Find the derivative of g(x)

First, we need to find the derivative of the function \( g(x) = x^3 - 4x + 6 \). The derivative \( g'(x) \) is obtained by applying the power rule to each term in the polynomial:- The derivative of \( x^3 \) is \( 3x^2 \).- The derivative of \( -4x \) is \( -4 \).- The derivative of the constant term \( 6 \) is \( 0 \).Thus, \( g'(x) = 3x^2 - 4 \).
02

Assign f'(x) equal to g'(x)

Given that \( f'(x) = g'(x) \), we have from Step 1:\[ f'(x) = 3x^2 - 4 \]
03

Integrate f'(x) to find f(x)

To find \( f(x) \), we integrate \( f'(x) \):\[ \int (3x^2 - 4) \, dx \]- The integral of \( 3x^2 \) is \( x^3 \).- The integral of \( -4 \) is \( -4x \).Thus, \( f(x) = x^3 - 4x + C \), where \( C \) is the integration constant.
04

Solve for the constant C using f(1)=2

We use the given condition \( f(1) = 2 \) to determine the constant \( C \).Substitute \( x = 1 \) into the equation \( f(x) = x^3 - 4x + C \):\[ f(1) = 1^3 - 4(1) + C = 2 \]Simplifying gives:\[ 1 - 4 + C = 2 \]\[ -3 + C = 2 \]Solving for \( C \):\[ C = 5 \]
05

Conclude f(x)

Now that we have the constant \( C \), we can write the function:\( f(x) = x^3 - 4x + 5 \).

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Key Concepts

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

The Derivative
In calculus, the derivative represents how a function changes as its input changes. For a given function, its derivative provides the rate of change or slope of the function at any point. In this case, we focus on the function \( g(x) = x^3 - 4x + 6 \). To find the derivative, \( g'(x) \), we apply the power rule to each term.
  • The derivative of \( x^3 \) is determined by multiplying 3 (the current power) by the coefficient (which is 1) and reducing the power by one, giving us \( 3x^2 \).
  • For the term \( -4x \), the derivative is simply the constant factor \(-4\) (since the power of \( x \) is 1).
  • The derivative of a constant, such as 6, is zero since a constant term does not change.
This yields the derivative \( g'(x) = 3x^2 - 4 \), which tells us how \( g(x) \) changes with \( x \).
Integration
Integration is the inverse process of differentiation. When we integrate, we're looking for a function that, when differentiated, gives us the original function's rate of change. In this exercise, we are given \( f'(x) = 3x^2 - 4 \) and need to determine \( f(x) \) by integrating.
  • The integral of \( 3x^2 \) can be found by increasing the power of \( x \) by one to get \( x^3 \), and dividing by the new power, resulting in \( x^3 \).
  • The integral of a constant, such as \( -4 \), is \( -4x \) because we essentially multiply by \( x^1/1 \).
Putting these together, the indefinite integral is \( f(x) = x^3 - 4x + C \), where \( C \) is the constant of integration, an essential inclusion in any indefinite integral as it accounts for any constant shift in the function.
Power Rule
The power rule is a cornerstone of calculus, making the process of finding derivatives straightforward by a specific formula. It states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule is not only useful but essential when dealing with polynomials, as it can be applied to each term independently.
  • In our original function \( g(x) = x^3 - 4x + 6 \), the power rule allows us to say the derivative of \( x^3 \) is \( 3x^2 \), simplifying what would otherwise be a complex process.
  • Furthermore, understanding the power rule facilitates ease in re-integrating polynomials to find functions from their derivatives, as seen when determining \( f(x) \).
By knowing and applying the power rule, we can efficiently move from functions, their derivatives, and back again through integration.
Constant of Integration
The constant of integration, \( C \), is an integral part of indefinite integration. Because derivatives of constant terms are zero, when we work backwards to find a function from its rate of change (its derivative), we must include an arbitrary constant. This "+C" accounts for any constant shift that occurred when differentiating.
  • In our case, once we found \( f(x) = x^3 - 4x + C \), it was crucial to identify \( C \) by using the condition \( f(1) = 2 \).
  • By substituting \( x = 1 \) and \( f(x) = 2 \) into this form, we found \( 1^3 - 4 \times 1 + C = 2 \), and hence solved for \( C = 5 \).
This constant ensures the function accurately represents the original problem constraints, highlighting why it's crucial in calculus calculations.

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