Problem 27 - Calculus an Applied Approach

Question

find the given value. $$ f(x)=x^{2}\left(3 x^{2}+3 x-4\right) \quad f^{\prime \prime \prime}(-2) $$

Step-by-Step Solution

Answer

The value of the third derivative of the function at \(x = -2\) is \(-48\).

Step 1: Find the First Derivative \(f'(x)\)

Start by differentiating the function \(f(x)=x^{2}(3x^{2}+3x-4)\). Using the product rule (which states that the derivative of two functions multiplied together is the derivative of the first times the second, plus the first times the derivative of the second), the first derivative \(f'(x)\) can be calculated as follows: \(f'(x)= 2x(3x^{2}+3x-4) + x^{2}(6x+3)\). Simplify this to get \(f'(x) = 6x^{3} + 6x^{2} - 8x + 6x^{2} + 3x^{2} = 6x^{3} + 12x^{2} - 8x\).

Step 2: Find the Second Derivative \(f''(x)\)

Next, differentiate \(f'(x)\) to get the second derivative \(f''(x)\). The second derivative can be calculated as: \(f''(x) = 18x^{2} + 24x - 8\).

Step 3: Find the Third Derivative \(f'''(x)\)

Now differentiate \(f''(x)\) to get the third derivative \(f'''(x)\). The third derivative can be calculated as: \(f'''(x) = 36x + 24\).

Step 4: Evaluate \(f'''(x)\) at \(x = -2\)

Finally, substitute \(x = -2\) into the equation for \(f'''(x)\) to find the value at this point. This gives: \( f'''(-2) = 36(-2) + 24 = -48\).

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus. It provides a way to differentiate two functions that are multiplied together. In other words, if you have a product of two functions, say \( u(x) \) and \( v(x) \), the Product Rule tells us how to find the derivative of this product.To apply the Product Rule, follow this simple formula:

• If \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This means you take the derivative of the first function, multiply it by the second function, then add this to the first function multiplied by the derivative of the second function.

In our exercise, we applied the Product Rule to differentiate \( f(x) = x^2(3x^2 + 3x - 4) \), resulting in the calculation:

• First derivative of \( x^2 \) is \( 2x \).
• Derivative of \( 3x^2 + 3x - 4 \) is \( 6x + 3 \).

By the Product Rule, the derivative \( f'(x) \) becomes \( 6x^3 + 12x^2 - 8x \).
Understanding the Product Rule is essential as it is used frequently in calculus for working with multiplicative functions.

First Derivative

The First Derivative of a function provides insight into its rate of change. Essentially, it tells us how quickly or slowly the function is increasing or decreasing at any point.Finding the first derivative, denoted as \( f'(x) \), involves taking the derivative of the original function. In our specific example, the function was \( f(x) = x^2(3x^2 + 3x - 4) \). We used the Product Rule to calculate this first derivative as:

• \( 2x(3x^2 + 3x - 4) + x^2(6x + 3) \)

After simplifying, this gives \( f'(x) = 6x^3 + 12x^2 - 8x \).

This expression is now much easier to interpret in terms of the function's behavior. Generally, when a derivative is positive, the function is increasing, while a negative derivative indicates a decreasing function.
Being able to compute and understand the first derivative is crucial for analyzing how functions behave.

Second Derivative

The Second Derivative provides additional information about the behavior of a function, specifically the acceleration or curvature of the graph. It is the derivative of the first derivative.By differentiating the first derivative \( f'(x) = 6x^3 + 12x^2 - 8x \), we obtain the second derivative. In our example:

• \( f''(x) = 18x^2 + 24x - 8 \)

This tells us about the concavity of the function's graph. If the second derivative is positive, the graph is concave upwards, resembling a U shape. If it is negative, the graph is concave downwards, resembling an upside-down U.

Understanding the second derivative is helpful for identifying important qualitative features of the original function's graph, such as inflection points where the curvature changes direction.

Third Derivative

The Third Derivative is a higher order derivative that gives more insights into the behavior of a function. It's particularly useful for understanding the jerk, or rate of change of acceleration.In our example problem, after finding the second derivative \( f''(x) = 18x^2 + 24x - 8 \), we calculated the third derivative by differentiating again:

• \( f'''(x) = 36x + 24 \)

The third derivative can reveal more subtle changes in the nature of the graph of the function. In practical terms, when evaluating this derivative at a specific point, like \( x = -2 \):

• Resulting in \( f'''(-2) = 36(-2) + 24 = -48 \)

This value allows us to assess how rapidly the second derivative is itself changing at this particular x-value.

In essence, working with higher order derivatives gives a comprehensive understanding of the subtleties inherent in the original function's behavior.