Problem 18 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative of the function at \(c = -3\) is 36.

Step 1: Identify the function and point

We are given the function \(f(x) = 2x^3 + 3x^2\) and need to find the derivative at the point \(c = -3\).

Step 2: Find the derivative of the function

Differentiate \(f(x) = 2x^3 + 3x^2\) to find the general form of the derivative. Using the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). Thus, the derivative \(f'(x) = 3 imes 2x^{2} + 2 imes 3x^{1}\). Simplifying gives: \(f'(x) = 6x^2 + 6x\).

Step 3: Substitute the point into the derivative

Now that we have \(f'(x) = 6x^2 + 6x\), substitute \(c = -3\) into the derivative function. Calculate: \(f'(-3) = 6(-3)^2 + 6(-3)\).

Step 4: Simplify the expression

Simplifying gives \(f'(-3) = 6 imes 9 + 6 imes (-3)\). Further simplify this to get \(f'(-3) = 54 - 18\).

Step 5: Final calculation

Perform the final arithmetic: \(f'(-3) = 36\).

Key Concepts

Differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function changes at any given point. When we differentiate a function, we are essentially looking for its derivative, which represents the slope of the function at that point. In our exercise, the goal was to compute the derivative of the function \(f(x) = 2x^3 + 3x^2\) at a specific point, \(c=-3\). By identifying the function and the specific point of interest, we set the stage for calculating the gradient or slope of the tangent line at that point.
The process of differentiation helps us understand how various quantities change in relation to one another. It's a fundamental tool for analyzing both linear and nonlinear functions, providing insights into the behavior of real-world systems. By differentiating, we can predict how small changes in one variable might influence another.

Power Rule

The power rule is a vital shortcut in differentiation that simplifies the process of finding derivatives for polynomial functions. It states that for any function of the form \(x^n\), the derivative is \(nx^{n-1}\). This rule is particularly useful as it eliminates the need for more complex operations like limits.
Applying the power rule to our example, we had \(f(x) = 2x^3 + 3x^2\). By using the power rule:

• For the term \(2x^3\), the derivative becomes \(3 \cdot 2x^{3-1} = 6x^2\).
• For \(3x^2\), the derivative is \(2 \cdot 3x^{2-1} = 6x\).

This gives us the derivative function \(f'(x) = 6x^2 + 6x\). The power rule makes computing derivatives for each term in a polynomial straightforward and efficient.

Function Analysis

Function analysis in calculus involves examining how a function behaves at specific intervals or points. This process typically includes determining its rate of change, as well as identifying critical points where the function's behavior might shift – such as maxima, minima, or points of inflection.
For the function \(f(x) = 2x^3 + 3x^2\), analyzing the derivative \(f'(x)\) and computing \(f'(-3)\) provide insights into the nature of function at \(x = -3\). By substituting \(x = -3\) into \(f'(x) = 6x^2 + 6x\), we calculated \(f'(-3) = 36\), indicating a positive slope at that point.
This result helps us deduce that the function is increasing at \(x = -3\). Function analysis like this is useful for sketching graphs, optimizing scenarios, and understanding the real-world meanings of mathematical relationships.

Calculus Problem Solving

Solving problems in calculus requires a structured approach that often involves decomposition into manageable steps. This methodology allows for breaking down complex problems into simpler parts, which can be solved systematically.
In this exercise, the problem-solving process included:

• Identifying the function and specific point of interest.
• Applying differentiation rules, such as the power rule, to find the general derivative.
• Substituting the given point into the derivative to find the slope.
• Simplifying the calculations to derive the final result.

By handling each element systematically, calculus problems become less daunting and more intuitive. This problem-solving skill is fundamental not just in mathematics, but in various fields where complex systems are analyzed and optimized.