Chapter 4: Problem 67
Consider the function \(f\) that is continuous on \([-1,1]\), such that \(f(0)=0\)
and
$$
f^{\prime}(x)=\left\\{\begin{array}{ll}
3 x^{2} & \text { for }-1
Short Answer
Expert verified
The formula for \( f(x) \) is \( f(x) = x^3 \) for \(-1 < x < 0\) and \( f(x) = 2e^x - 2 \) for \(0 < x < 1\).
Step by step solution
01
Integrate to Find Antiderivatives
First, we'll find a formula for the function by integrating the given derivative piece-wise. Since \( f'(x) = 3x^2 \) for \(-1 < x < 0\), we integrate this:\[ F(x) = \int 3x^2 \, dx = x^3 + C_1 \]For \( 0 < x < 1 \), since \( f'(x) = 2e^x \), we integrate this:\[ G(x) = \int 2e^x \, dx = 2e^x + C_2 \].
02
Use Boundary Condition at x = 0
We know that \( f(0) = 0 \). Both expressions for \( f(x) \) must equal 0 at the boundary point \( x = 0 \) to ensure continuity. For the left-hand side (LHS), \( f(0^-) \), we substitute \( x = 0 \) in:\[ F(0) = 0^3 + C_1 = 0 \Rightarrow C_1 = 0 \]For the right-hand side (RHS), \( f(0^+) \), we substitute \( x = 0 \) in:\[ G(0) = 2e^0 + C_2 = 0 \Rightarrow C_2 = -2 \].
03
Write the Piecewise Function
Using the integration constants found, we write the piecewise function for \( f(x) \):\[ f(x) = \begin{cases} x^3, & \text{if } -1 < x < 0 \ 2e^x - 2, & \text{if } 0 < x < 1 \end{cases} \]
04
Ensure Continuity at x = 0
To ensure \( f(x) \) is continuous at \( x = 0 \), confirm that both pieces give the same value at the boundary. From our conditions, \( f(0^-) = 0^3 = 0 \) and \( f(0^+) = 2e^0 - 2 = 0 \). The function is continuous at \( x = 0 \).
05
Sketch the Graph
Now we plot each segment of the piecewise function. For \(-1 < x < 0\), plot \( f(x) = x^3 \), which is a cubic function smoothly rising from below -1 to 0. For \(0 < x < 1\), plot \( f(x) = 2e^x - 2 \), which is an exponential curve shifted down by 2 units.
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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 that involves finding the antiderivative of a function. This process helps us determine the original function when its derivative is given. In our exercise, we deal with a piecewise function, where each section of the derivative, \( f'(x) \), needs a separate integration process.
For
For
- \(-1 < x < 0\), we are given \( f'(x) = 3x^2 \). Integrating this provides us with one part of the original function: \( F(x) = x^3 + C_1 \).
- For \(0 < x < 1\), \( f'(x) = 2e^x \) is given. Integration here leads us to another part of the function: \( G(x) = 2e^x + C_2 \).
Continuity
Continuity in mathematics means that a function is unbroken and has no gaps over its specified domain. For a piecewise function like ours, achieving continuity requires that the pieces of the function connect smoothly without jumps or breaks.
A pivotal part of ensuring continuity is matching the values of the function at the points where the pieces meet. In our exercise:
A pivotal part of ensuring continuity is matching the values of the function at the points where the pieces meet. In our exercise:
- The left side of the function, \( f(x) = x^3 \), and the right side, \( f(x) = 2e^x - 2 \), both connect at \( x = 0 \).
- By verifying that both expressions equal zero at this point (\( f(0^-) = f(0^+) = 0 \)), we ensure the function is continuous at \( x = 0 \).
Antiderivatives
An antiderivative is a function that reverses differentiation. It essentially helps us retrieve the original function from its derivative. When dealing with piecewise-defined functions, finding the antiderivative for each section is key to reconstructing the entire function.
For our exercise:
For our exercise:
- The antiderivative of \( 3x^2 \) for the interval \(-1 < x < 0\) is \( x^3 + C_1 \).
- The antiderivative of \( 2e^x \) for \( 0 < x < 1 \) is \( 2e^x + C_2 \).
Boundary Conditions
Boundary conditions are essential to solving problems involving differential equations or piecewise functions. These conditions allow us to tie different parts of the function together, ensuring a smooth and coherent definition across boundaries.
In this exercise, one such boundary condition was provided at \( x = 0 \), where \( f(0) = 0 \). This was critical for pinpointing the constants of integration and ensuring continuity:
In this exercise, one such boundary condition was provided at \( x = 0 \), where \( f(0) = 0 \). This was critical for pinpointing the constants of integration and ensuring continuity:
- From the left, substituting into \( x^3 + C_1 = 0 \) at \( x = 0 \) yielded \( C_1 = 0 \).
- From the right, \( 2e^0 + C_2 = 0 \) at \( x = 0 \) provided \( C_2 = -2 \).