Question

If \(f^{\prime}(x)=3 x+2,\) find the slope of the line tangent to the curve \(y=f(x)\) at \(x=1,2,\) and 3

Short answer

Answer: At \(x = 1\), the slope is \(5\). At \(x = 2\), the slope is \(8\). At \(x = 3\), the slope is \(11\).

Step-by-step solution

Find the derivative of the function

We are given the derivative of the function as \(f'(x) = 3x + 2\)

Evaluate the derivative at x = 1

To find the slope of the tangent line at x = 1, substitute x = 1 into the derivative: \(f'(1) = 3(1) + 2 = 3 + 2 = 5\)

Evaluate the derivative at x = 2

To find the slope of the tangent line at x = 2, substitute x = 2 into the derivative: \(f'(2) = 3(2) + 2 = 6 + 2 = 8\)

Evaluate the derivative at x = 3

To find the slope of the tangent line at x = 3, substitute x = 3 into the derivative: \(f'(3) = 3(3) + 2 = 9 + 2 = 11\) The slopes of the line tangent to the curve \(y=f(x)\) at the given points are: At \(x=1\), the slope is \(5\). At \(x=2\), the slope is \(8\). At \(x=3\), the slope is \(11\).

Key concepts

Derivative

In calculus, the derivative is foundational for understanding rates of change. It essentially measures how a function's output value changes in response to alterations in its input value. When we say we are taking the derivative of a function, symbolically represented as \( f'(x) \) or \( \frac{df}{dx} \), we're looking for a formula that tells us the slope of the tangent line at any point along the curve that represents the function.

For instance, in the exercise given, \( f'(x) = 3x + 2 \) is the derivative of some function \( f(x) \). This equation represents the slope of the tangent line to the curve at any value of \( x \). Thus, understanding derivatives is crucial in solving problems related to rates of change, such as velocity in physics or optimizing functions in economics.

Slope of Tangent

The slope of the tangent line to a curve at a particular point provides a precise measure of the steepness of the curve at that exact location. For any differentiable function, the slope of the tangent at any point is precisely the value of the derivative of the function at that point. In graphical terms, if you were to zoom in on the curve at a point where it's smooth and differentiable, the curve would begin to look straighter, and its slope at that point is what we're interested in.

For example, when we were asked to find the slope of the line tangent to the curve \( y=f(x) \) at \( x=1,2, \) and 3, we simply substituted these values into the given derivative formula to find the respective slopes. This concept allows us to understand the behavior of functions and is a vital concept in fields like engineering and physics.

Evaluating Derivatives

Evaluating derivatives is a practical process. Once we have the derivative function, as in our exercise where \( f'(x) = 3x + 2 \), we can find out the rate of change at specific points. This is done by plugging in the values of the independent variable—\( x \) in this case—into the derivative function.

Here's a step-by-step approach to evaluating the derivative, which can reinforce our exercise examples:

• Identify the derivative function (given or calculated).
• Substitute the value of \( x \) into the derivative equation.
• Simplify to solve for the derivative's value at that \( x \).

These values tell us the instantaneous slope of the original function at the given values of \( x \).

Calculus Problem Solving

Calculus problem solving involves identifying the necessary calculus principles and applying them correctly to the given problem. In the context of our exercise, recognizing that the slope of the tangent line at a point can be determined by evaluating the derivative at that point was the first step. Afterward, by systematically substituting the given \( x \) values—1, 2, and 3—into the derivative and calculating the result, we successfully solved the problem.

To effectively tackle calculus problems, it's important to:

• Understand the underlying principles and formulas.
• Methodically work through each step.
• Double-check calculations to avoid errors.

Calculus is not just theoretical; it's a powerful tool used to solve real-world problems in various fields such as physics, engineering, economics, and even medicine.