Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 5 through 8, 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) \(f^{\prime}(x)=3\) (b) \(f^{\prime}(x)=\pi\)

Step-by-Step Solution

Answer

The function for part (a) is \(f(x) = 3x + 0\) or simply \(f(x) = 3x\). The function for part (b) is \(f(x) = \pi x + 2\).

Step 1 Find Antiderivative for (a)

The antiderivative \(f(x)\) of the derivative \(f^{\prime}(x)=3\) is calculated simply by taking the integral of the derivative. That integral would result in a linear function \(f(x) = 3x + C\), where C is the constant of integration.

Step 2 Apply Constraint for (a)

The constraint given is that \(f(x)\) passes through the origin, or the point (0,0). This means that when x = 0, y (or \(f(x)\)) must also equal 0. Apply this to your antiderivative to solve for C. \(f(0) = 3*0 + C = 0\), so C = 0.

Step 3 Find Antiderivative for (b)

The antiderivative \(f(x)\) of the derivative \(f^{\prime} (x)=\pi\) is calculated similar to step 1 by taking the integral of the derivative. It results in another linear function \(f(x) = \pi x + C\), where C is the constant of integration.

Step 4 Apply Constraint for (b)

The constraint given in part (b) is that \(f(x)\) passes through the point (0,2). This means that when x = 0, \(f(x)\) must equal 2. Apply this condition to your antiderivative to solve for C. \(f(0) = \pi*0 + C = 2\), so C = 2.

Key Concepts

Integrals in Calculus

Understanding the concept of integrals in calculus is a foundation stone for delving into the continuous aspects of mathematics. An integral can be seen as the opposite operation to taking a derivative, and it essentially represents the area under a curve defined by a function on a graph.

For instance, if we're given a constant function like f'(x) = 3, its integral will give us the original function that has a slope of 3 across its graph. It represents a 'summing up' process—or in this case, the accumulation of area under the straight line. In school, you've perhaps filled shapes to calculate area; think of integration as a way to fill under the curve, even when that curve can change.

This 'summing up' is also the antidote to the rate of change provided by derivatives, bridging the gap between discrete and continuous quantities.

Constant of Integration

When performing antiderivative calculations, introducing the constant of integration, typically denoted as C, is crucial. This constant represents any constant value that could have been in the original function before it was differentiated. Because the derivative of any constant is zero, when we integrate, we must include this C to account for all possible original functions.

For example, when we integrated f'(x) = 3 to find f(x) = 3x + C, we have an infinite number of linear functions that differ just by their y-intercept due to the constant C. The additional information, like the function passing through the origin, allows us to solve for C. It's like a detective finding a clue; out of many suspects (possible constants), only one fits perfectly with our 'scene of the crime' (the given point on the graph).

Derivative Function

The derivative function is a powerful tool in calculus that tells us about the rate at which things change. The derivative of a function at any point gives the slope of the tangent line to the curve at that point, representing how fast the function's value is changing with respect to a change in its input value.

Take any routine activity, like driving—the speedometer shows your instantaneous rate of change of position, which is actually a real-life derivative! Similarly, f'(x) informs us of how f(x) behaves instantaneously. For the exercise we're examining, f'(x) = 3 informs us that for every one unit increase in x, f(x) increases by three units— a constant rate, corresponding to moving at a constant speed in our driving analogy.