Problem 53 - Calculus: Early Transcendentals

Question

Use the definition of a derivative to find \( f'(x) \) and \( f''(x) \). Then graph \( f \), \( f' \), and \( f'' \) on a common screen and check to see if your answers are reasonable. \( f(x) = 3x^2 + 2x + 1 \)

Step-by-Step Solution

Answer

\( f'(x) = 6x + 2; \ f''(x) = 6 \). Graphing shows the quadratic function and its derivatives as expected.

Step 1: Understanding the Definition of Derivative

The derivative of a function at a point is the slope of the tangent to the graph of the function at that point. It can be calculated using the limit definition: \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \). We will use this formula to find the first derivative of \( f(x) = 3x^2 + 2x + 1 \).

Step 2: Apply the Definition to Find f'(x)

Start by writing the function with \( x+h \): \( f(x+h) = 3(x+h)^2 + 2(x+h) + 1 \). Expand \( (x+h)^2 = x^2 + 2xh + h^2 \) and substitute back. So, \( f(x+h) = 3(x^2 + 2xh + h^2) + 2x + 2h + 1 \). Simplifying, we get \( 3x^2 + 6xh + 3h^2 + 2x + 2h + 1 \). Now, calculate the difference: \( f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 2x + 2h + 1) - (3x^2 + 2x + 1) \). The result is \( 6xh + 3h^2 + 2h \).

Step 3: Take the Limit to Find f'(x)

Using the expression \( 6xh + 3h^2 + 2h \), calculate the derivative using the limit: \( f'(x) = \lim_{{h \to 0}} \frac{6xh + 3h^2 + 2h}{h} \). Factor out \( h \) from the numerator: \( f'(x) = \lim_{{h \to 0}} (6x + 3h + 2) \). As \( h \to 0 \), this simplifies to \( 6x + 2 \). This is the first derivative, \( f'(x) = 6x + 2 \).

Step 4: Find the Second Derivative f''(x)

To find the second derivative \( f''(x) \), differentiate \( f'(x) = 6x + 2 \) with respect to \( x \). The derivative of \( 6x \) is 6 and the derivative of a constant, \( 2 \), is 0. Thus, \( f''(x) = 6 \).

Step 5: Graph f, f', and f''

To understand how \( f(x), f'(x), \) and \( f''(x) \) behave visually, plot all three. \( f(x) = 3x^2 + 2x + 1 \) is a parabola opening upwards. \( f'(x) = 6x + 2 \) is a linear function with a constant slope. \( f''(x) = 6 \) is a horizontal line, indicating a constant curvature.

Key Concepts

Limit Definition of Derivative

Understanding the limit definition of the derivative is crucial in calculus. This concept helps us find the derivative, or the instantaneous rate of change, of a function at any given point. The formal definition states that if you have a function \( f(x) \), its derivative is given by \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \].
This equation is about finding the limit of the average rate of change as the interval \( h \) approaches zero.This method tells us the slope of the tangent line at each point \( x \) on the graph of \( f(x) \).

• Start by calculating \( f(x+h) \).
• Subtract \( f(x) \) to get the difference.
• Divide the result by \( h \).
• Find the limit as \( h \to 0 \).

Applying these steps to our function, \( f(x) = 3x^2 + 2x + 1 \), results in \( f'(x) = 6x + 2 \). This will tell us not just how the function grows, but also its curvature at every point.

Second Derivative

The second derivative is a powerful tool that gives us insight into the concavity of a function. Given a function's first derivative, \( f'(x) \), the process to find the second derivative, denoted as \( f''(x) \), involves taking the derivative of the first derivative.
For the function \( f(x) = 3x^2 + 2x + 1 \), we found that the first derivative is \( f'(x) = 6x + 2 \). Now, by differentiating this expression, we get \( f''(x) = 6 \).

• The derivative of \( 6x \) is 6, as we are differentiating with respect to \( x \).
• Mathematically, the derivative of a constant (2) is 0.

A constant second derivative tells us that the original function has a uniform concavity; this is why \( f(x) \) is a parabola, always opening upwards, with no change in curvature.

Graphing Functions

Graphing is an essential skill in visualizing how derivatives affect the shape and behavior of functions. Plotting \( f(x) \), \( f'(x) \), and \( f''(x) \) on a single graph offers a deeper understanding of their relationships.
Let's consider the functions derived from \( f(x) = 3x^2 + 2x + 1 \):

• \( f(x) \) graph: This is a parabola, clearly demonstrating how the function behaves symmetrically around its vertex.
• \( f'(x) \) graph: It is a straight line, illustrating changes in slope for \( f(x) \). The slope is positive, indicating that \( f(x) \) increases, steeply as \( x \) moves away from the vertex in both directions.
• \( f''(x) \) graph: Being a constant equals a horizontal line. This confirms that the rate of change of the slope (or the curvature) of \( f(x) \) is consistent at 6.

Visualizing these graphs helps to quickly confirm the mathematical conclusions about the growth, stability, and shape of \( f(x) \).