Problem 55 - Calculus: Early Transcendentals

Question

If \( f(x) = 2x^2 - x^3 \), find \( f'(x) \), \( f''(x) \), \( f'''(x) \), and \( f^{(4)}(x) \). Graph \( f \), \( f' \), \( f'' \), and \( f''' \) on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?

Step-by-Step Solution

Answer

The derivatives are consistent with their geometric interpretations: diminishing degrees of curves and constant changes.

Step 1: Find First Derivative

To find the first derivative of the function \( f(x) = 2x^2 - x^3 \), apply the power rule. The power rule states that if \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \). So:\[f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(x^3)\]\[f'(x) = 2 \cdot 2x^{2-1} - 3 \cdot x^{3-1}\]\[f'(x) = 4x - 3x^2\].

Step 2: Find Second Derivative

We will now find the derivative of \( f'(x) = 4x - 3x^2 \) to get the second derivative \( f''(x) \). Again using the power rule:\[f''(x) = \frac{d}{dx}(4x) - \frac{d}{dx}(3x^2)\]\[f''(x) = 4 \cdot 1x^{1-1} - 3 \cdot 2x^{2-1}\]\[f''(x) = 4 - 6x\].

Step 3: Find Third Derivative

Next, find the derivative of \( f''(x) = 4 - 6x \) to find \( f'''(x) \). This gives:\[f'''(x) = \frac{d}{dx}(4) - \frac{d}{dx}(6x)\]\[f'''(x) = 0 - 6 \cdot 1x^{1-1}\]\[f'''(x) = -6\].

Step 4: Find Fourth Derivative

Finally, find the derivative of \( f'''(x) = -6 \) to find \( f^{(4)}(x) \):\[f^{(4)}(x) = \frac{d}{dx}(-6)\]\[f^{(4)}(x) = 0\].

Step 5: Graph the Functions

Now, graph the original function \( f(x) = 2x^2 - x^3 \) along with its derivatives \( f'(x) = 4x - 3x^2 \), \( f''(x) = 4 - 6x \), and \( f'''(x) = -6 \) on a common graph using software or graphing tools. Interpretation:- \( f(x) \) is a cubic polynomial, indicating a curve that might have local maxima or minima.- \( f'(x) \) is a quadratic function, represented by a parabola; it shows the slope changes.- \( f''(x) \) is linear, describing the rate of change of the slope of \( f \).- The constant \( f'''(x) \) indicates a uniform curvature rate, and \( f^{(4)}(x) \) being zero confirms no further changes in curvature beyond all previous derivatives.

Key Concepts

Power Rule

The power rule is an essential tool in calculus for finding derivatives of polynomial functions. It simplifies the process of derivative calculation and is particularly useful for functions expressed as powers of a variable. In its simplest form, if you have a function like \( f(x) = x^n \), the derivative of this function, expressed as \( f'(x) \), can be found using:

• Multiply the power of \( x \) by its coefficient.
• Reduce the power by one.

So if \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).
This rule allows us to quickly find the rate of change of polynomial functions. For example, in our exercise, to find the derivative of \( f(x) = 2x^2 - x^3 \), we apply the power rule as follows:

• The derivative of \( 2x^2 \) is \( 2 \cdot 2x^{2-1} = 4x \).
• The derivative of \( -x^3 \) is \( -3 \cdot x^{3-1} = -3x^2 \).

This results in the first derivative \( f'(x) = 4x - 3x^2 \). Understanding and applying the power rule is crucial for efficiently working with polynomial derivatives.

Cubic Polynomial

A cubic polynomial is a type of polynomial with a degree of three. This means the highest exponent of the variable is three. These functions have the general form of \( ax^3 + bx^2 + cx + d \), where \( a \), \( b \), and \( c \) are coefficients, and \( d \) is the constant term. The shape of a cubic polynomial can be quite dynamic, often featuring curves with local maxima and minima.
For the function \( f(x) = 2x^2 - x^3 \) given in our exercise, it's represented as \( -x^3 + 2x^2 + 0x + 0 \).
One characteristic of cubic polynomials is that they can have turning points. These points are significant because they indicate where the function changes from increasing to decreasing, or vice versa. To find these points, one typically examines the first derivative \( f'(x) \) to determine the slopes of the tangent at various points on the curve. Since the first derivative \( f'(x) = 4x - 3x^2 \) is a quadratic equation, solving it yields the critical points of the original cubic polynomial. Thus, understanding cubic polynomials helps in predicting the behavior and graph of the function.

Graphing Derivatives

Graphing derivatives involves visualizing the behavior of a function and its subsequent derivatives on a graph. This process helps us understand how the function's slope and curvature change across its domain.
Let's consider our function and its derivatives from the exercise, where:

• \( f(x) = 2x^2 - x^3 \)
• \( f'(x) = 4x - 3x^2 \)
• \( f''(x) = 4 - 6x \)
• \( f'''(x) = -6 \)

When graphing, \( f(x) \), the original cubic polynomial, is expected to have a curve with possible peaks and valleys. \( f'(x) \) is a quadratic function, typically appearing as a parabola on the graph. It shows us where the slope of \( f(x) \) is positive, negative, or zero (flat).
\( f''(x) \) is linear, indicating where \( f'(x) \) is increasing or decreasing. Changes in the slope's rate are shown here. Finally, \( f'''(x) \) being a constant suggests the rate of slope change is uniform.
Visualizing these on a common graph provides a comprehensive picture of the function's behavior. The steady slope in \( f'''(x) \) reflects a constant rate of change of the slope, while the zeros in \( f''(x) \) give information about potential inflection points on \( f(x) \). This graphing method not only verifies the mathematical calculations but also offers intuitive insight into the function's dynamics.