Problem 20 - Calculus Single Variable

Question

A function \(f\) and a point \(c\) are given. Calculate \(f^{\prime}(c)\). $$ f(x)=5 x^{3}+4 x^{2}+3 x+2 \quad c=-2 $$

Step-by-Step Solution

Answer

The derivative of the function at \( c = -2 \) is 47.

Step 1: Identify the Function and Point

We are given the function \( f(x) = 5x^3 + 4x^2 + 3x + 2 \) and we need to find the derivative at the point \( c = -2 \). This means we are looking for \( f'(c) \) where \( c = -2 \).

Step 2: Find Derivative of the Function

To find \( f'(x) \), differentiate \( f(x) = 5x^3 + 4x^2 + 3x + 2 \) term by term. The derivative \( f'(x) \) is found using rules of differentiation. The power rule states that if \( f(x) = ax^n \), then \( f'(x) = anx^{n-1} \).

Step 3: Apply the Power Rule

Using the power rule on each term: the derivative of \( 5x^3 \) is \( 15x^2 \), the derivative of \( 4x^2 \) is \( 8x \), and the derivative of \( 3x \) is \( 3 \). The derivative of a constant, \( 2 \), is \( 0 \). Thus, \( f'(x) = 15x^2 + 8x + 3 \).

Step 4: Evaluate the Derivative at the Point

Now substitute \( x = -2 \) into \( f'(x) = 15x^2 + 8x + 3 \). Calculate \( f'(-2) = 15(-2)^2 + 8(-2) + 3 \). First, compute each term: \( 15(-2)^2 = 15 imes 4 = 60 \), \( 8(-2) = -16 \), and \( 3 = 3 \).

Step 5: Calculate and Simplify

Add the results from the previous step: \( f'(-2) = 60 - 16 + 3 \). Simplify this to get \( f'(-2) = 47 \).

Key Concepts

Power Rule

The power rule is a fundamental component of differentiation, making the process of finding derivatives much more straightforward. It applies to any term in the form of \( ax^n \), where \( a \) is a constant and \( n \) is a positive integer. The rule specifies that the derivative of \( ax^n \) is \( anx^{n-1} \). This means:

• Bring down the power \( n \) as a coefficient.
• Multiply it by the existing coefficient \( a \).
• Reduce the power by one to get the new exponent.

For example, if we have a term \( 5x^3 \), using the power rule gives us \( 15x^2 \):- Multiply the coefficient 5 by 3 (the exponent), yielding the new coefficient 15.- Decrease the exponent 3 by 1 to get 2. Thus, the derivative is \( 15x^2 \).
This simple rule allows us to quickly find the derivative of polynomial terms, aiding in the broader process of differentiation.

Differentiation

Differentiation is the process used in calculus to determine the derivative of a function. The derivative itself is a measure of how the function changes as its input changes. It provides the slope of the function at any given point, which is crucial for analyzing the function's behavior.
The process involves applying differentiation rules, such as the power rule, to each term of a function individually and then summing the results. For polynomial functions, this means:- Differentiating each term separately.- Combining these to form the complete derivative function.
For instance, if we look at the function \( f(x) = 5x^3 + 4x^2 + 3x + 2 \), each term is derived separately:

• \( 5x^3 \) becomes \( 15x^2 \).
• \( 4x^2 \) becomes \( 8x \).
• \( 3x \) becomes \( 3 \).
• The constant \( 2 \) simply becomes \( 0 \) because the slope of a constant is zero.

Finally, combining these results gives the derivative \( f'(x) = 15x^2 + 8x + 3 \). Differentiation allows us to analyze and predict the function's behavior at different points, highlighting key features such as increasing or decreasing intervals and curvature.

Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of powers of a variable. They are of the form \( a_nx^n + a_{n-1}x^{n-1} + \, \ldots \, + a_1x + a_0 \), where each \( a_i \) is a constant coefficient and \( n \) is a non-negative integer representing the degree of the polynomial. These functions are notable for their smooth, continuous graphs which can be linear, quadratic, cubic, or higher degrees, depending on the highest power present.
The process of finding derivatives for polynomial functions is simplified by working with each term separately using the power rule. Polynomial functions provide a flexible framework for modeling a wide variety of real-world situations. Their derivatives, obtained through differentiation, help in understanding:

• The rate at which quantities change.
• The slopes of curves at various points, indicating where they rise or fall.
• Critical points where behavior changes, such as turning points.

For example, the polynomial function \( f(x) = 5x^3 + 4x^2 + 3x + 2 \) possesses terms of varying degrees that describe complex curves. Its derivative \( f'(x) = 15x^2 + 8x + 3 \) reveals how these curves grow or diminish at different intervals, helping to pinpoint behavior changes.