Problem 37
Question
Find the function with the given derivative whose graph passes through the point \(P.\) \begin{equation}f^{\prime}(x)=2 x-1, \quad P(0,0)\end{equation}
Step-by-Step Solution
Verified Answer
The function is \( f(x) = x^2 - x \).
1Step 1: Understanding the Problem
We need to find the original function, \( f(x) \), given its derivative \( f'(x) = 2x - 1 \). Additionally, the function should pass through the point \( P(0, 0) \).
2Step 2: Integrating the Derivative
To find the original function, we integrate its derivative. The integral of \( f'(x) = 2x - 1 \) is\[\int (2x - 1) \, dx = x^2 - x + C,\]where \( C \) is the constant of integration.
3Step 3: Using the Initial Condition
We use the point \( P(0, 0) \) to find the value of \( C \). Substitute \( x = 0 \) and \( f(x) = 0 \) into the equation:\[0 = 0^2 - 0 + C \ C = 0.\]Thus, the function is \( f(x) = x^2 - x \).
4Step 4: Verifying the Solution
We verify by substituting back. The derivative is \( f'(x) = 2x - 1 \), which matches the given derivative. The point \( (0, 0) \) is on the graph of \( f(x) = x^2 - x \). Everything checks out.
Key Concepts
DerivativeIntegrationFunction Graphs
Derivative
In calculus, the derivative represents the rate at which a function is changing at any given point. It provides a way to understand how the function's value changes as its input, or variable, changes.
The derivative of a function, often denoted as \( f'(x) \), gives us a new function that tells us the slope of the original function at every point. In simpler terms, the derivative describes how steep the curve is at any location. For example, the given derivative \( f'(x) = 2x - 1 \) tells us that the rate of change of the function is linearly dependent on \( x \).
**Key Points to Remember:**
The derivative of a function, often denoted as \( f'(x) \), gives us a new function that tells us the slope of the original function at every point. In simpler terms, the derivative describes how steep the curve is at any location. For example, the given derivative \( f'(x) = 2x - 1 \) tells us that the rate of change of the function is linearly dependent on \( x \).
**Key Points to Remember:**
- The derivative is the ‘instantaneous rate of change’ of the function.
- Knowing the derivative allows us to find tangent lines and understand the behavior of graphs.
- Graphically, the derivative function indicates the slope of the function at various points.
Integration
Integration is essentially the reverse process of differentiation. It's about finding the original function from its derivative. When we integrate a derivative function, we retrieve the original function up to an unknown constant. This process is termed as anti-differentiation or simply integration.
For the problem at hand, integrating \( f'(x) = 2x - 1 \) results in:\[\int (2x - 1) \, dx = x^2 - x + C,\]where \( C \) is a constant known as the constant of integration. This constant \( C \) emerges because integration can generate infinitely many functions differing only by a constant.
**Important Aspects of Integration:**
For the problem at hand, integrating \( f'(x) = 2x - 1 \) results in:\[\int (2x - 1) \, dx = x^2 - x + C,\]where \( C \) is a constant known as the constant of integration. This constant \( C \) emerges because integration can generate infinitely many functions differing only by a constant.
**Important Aspects of Integration:**
- Integration helps in finding the general formula of the function from its known derivative.
- The constant \( C \) needs to be determined by using additional information like specific points on the function's graph.
- Graphically, integrating provides the accumulated area under the curve of the derivative.
Function Graphs
The graph of a function provides a visual representation of its behavior. When discussing derivatives and integrals, function graphs are immensely helpful.
To graph a function like \( f(x) = x^2 - x \), it's essential to understand its key features, such as roots, turning points, and asymptotic behavior. For the specific function in the exercise, the graph will pass through the origin \( (0, 0) \), as shown by substituting \( x = 0 \) in \( f(x) \). Another point, derived from the exercise, highlights the function crossing the x-axis wherever the function equals zero. Solving \( x^2 - x = 0 \) gives roots at \( x = 0 \) and \( x = 1 \). **Graph Features:**
To graph a function like \( f(x) = x^2 - x \), it's essential to understand its key features, such as roots, turning points, and asymptotic behavior. For the specific function in the exercise, the graph will pass through the origin \( (0, 0) \), as shown by substituting \( x = 0 \) in \( f(x) \). Another point, derived from the exercise, highlights the function crossing the x-axis wherever the function equals zero. Solving \( x^2 - x = 0 \) gives roots at \( x = 0 \) and \( x = 1 \). **Graph Features:**
- Roots: Points where the function crosses the x-axis.
- Turning Points: Locations where the function changes direction, found by solving \( f'(x) = 0 \).
- Slope: Determined by the derivative, \( f'(x) = 2x - 1 \), showing whether the function is increasing or decreasing.
Other exercises in this chapter
Problem 37
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