Problem 35 - Essential Calculus Early Transcendentals

Question

Given that the graph of \(f\) passes through the point \((1,6)\) and that the slope of its tangent line at \((x, f(x))\) is \(2 x+1\) find \(f(2)\) .

Step-by-Step Solution

Answer

\( f(2) = 10 \)

Step 1: Understand the Problem

We are given that the slope of the tangent line to the function \( f(x) \) is represented by the derivative \( f'(x) = 2x + 1 \). We also know that the function passes through the point \((1,6)\). We need to find the function \( f(x) \) and then evaluate \( f(2) \).

Step 2: Find the Antiderivative

Since \( f'(x) = 2x + 1 \), we need to find an antiderivative of this expression to determine \( f(x) \). The antiderivative of \( 2x \) is \( x^2 \), and the antiderivative of \( 1 \) is \( x \). Therefore, the general form of \( f(x) \) is \( f(x) = x^2 + x + C \), where \( C \) is the constant of integration that we need to determine.

Step 3: Use Given Point to Find \(C\)

We know that the function passes through the point \((1, 6)\). Substitute \( x = 1 \) and \( f(1) = 6 \) into the equation \( f(x) = x^2 + x + C \) to solve for \( C \).\[6 = 1^2 + 1 + C \6 = 1 + 1 + C \6 = 2 + C \C = 4\]Therefore, the specific function is \( f(x) = x^2 + x + 4 \).

Step 4: Evaluate \( f(2) \)

Now that we have the function \( f(x) = x^2 + x + 4 \), substitute \( x = 2 \) to find \( f(2) \).\[f(2) = 2^2 + 2 + 4 \f(2) = 4 + 2 + 4 = 10\]

Step 5: Solution Summary

After finding the expression for \( f(x) \) and evaluating it at \( x = 2 \), we determined that \( f(2) = 10 \).

Key Concepts

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function we started with. So, if you're looking at a function's derivative and want to reverse the process, you're finding the antiderivative.

In our problem, we started with the derivative of a function, given by the slope of the tangent line, which is \( f'(x) = 2x + 1 \). To find the original function \( f(x) \), we need to identify the antiderivative of \( 2x + 1 \). This means we're searching for a function whose derivative results in \( 2x + 1 \).

• The antiderivative of \( 2x \) is \( x^2 \).
• The antiderivative of \( 1 \) is \( x \).

Putting it together, the antiderivative is \( f(x) = x^2 + x + C \), where \( C \) is the constant of integration. This constant \( C \) accounts for any vertical shift in our function family, ensuring we encompass all possible original functions from which \( 2x + 1 \) could derive. This process establishes the potential forms \( f(x) \) could take.

Tangent Line

A tangent line to a curve at a given point is a straight line that lightly touches the curve at that point. It represents the instantaneous rate of change or slope of the curve at that specific point.

In this exercise, we dealt with the slope of the tangent line described by \( f'(x) = 2x + 1 \). This expression isn't just a single slope; it varies according to the value of \( x \). This variability means every point \( (x, f(x)) \) on our function \( f(x) \) has its unique tangent line slope based on the formula \( 2x + 1 \).

• This powerful concept helps in understanding how functions behave locally.
• At \( x = 1 \), the slope of the tangent is \( 2(1) + 1 = 3 \).
• This helps local approximation of the function, giving insights into its direction and steepness.

Thus, the tangent line is central to calculus as it uncovers not just how functions might be trending at a point, but also is a foundation for derivative concepts.

Constant of Integration

When you find an antiderivative of a function, as we did here, you come across a concept called the "constant of integration." This is represented by the symbol \( C \), and it represents an infinite number of parallels that could be drawn from the pattern of the function's slope.

Let's dive in a little deeper:

• As seen, the antiderivative of \( 2x + 1 \) is \( f(x) = x^2 + x + C \).
• Without additional information, we can't specifically determine \( C \); initially, it's undefined.
• To pinpoint \( C \), we use any known point on the function, such as \( (1, 6) \) in this exercise, enabling us to solve for \( C \) using \( 6 = 1^2 + 1 + C \), giving \( C = 4 \).

Therefore, the constant of integration is crucial because it defines the exact path the graph of the function takes. Without it, our function is just a family of curves, and \( C \) serves as the identifier for our specific function within this family.