Problem 3 - Calculus with Concepts in Calculus

Question

Determine the values of \(c\) at which \(f^{\prime}\) changes from positive to negative, or from negative to positive. $$ f(x)=2 x^{4}-4 x^{2}+3 $$

Step-by-Step Solution

Answer

The derivative changes sign at \( x = -1, 0, 1 \).

Step 1: Find the derivative of the function

To find where the derivative changes sign, we first need to find the derivative of the function. The function is given as: \[ f(x) = 2x^4 - 4x^2 + 3 \]The derivative, \( f'(x) \), is found using the power rule: \[ f'(x) = 8x^3 - 8x \]

Step 2: Set the derivative equal to zero

To find the critical points, where the sign of the derivative might change, set the derivative equal to zero:\[ 8x^3 - 8x = 0 \]

Step 3: Solve for critical points

Factor the equation to solve for \(x\):\[ 8x(x^2 - 1) = 0 \]This gives us:- \( 8x = 0 \) which yields \( x = 0 \)- \( x^2 - 1 = 0 \) which factors into \( (x - 1)(x + 1) = 0 \), yielding \( x = 1 \) and \( x = -1 \)Thus, the critical points are \( x = -1, 0, 1 \).

Step 4: Use a sign chart to determine sign changes

To determine where the derivative changes sign, evaluate \( f'(x) \) in the intervals around the critical points: - In the interval \((-\infty, -1)\), pick \(x = -2\): \(f'(-2) = 8(-2)^3 - 8(-2) = -64 + 16 = -48\) (negative)- In the interval \((-1, 0)\), pick \(x = -0.5\): \(f'(-0.5) = 8(-0.5)^3 - 8(-0.5) = -1 + 4 = 3\) (positive)- In the interval \((0, 1)\), pick \(x = 0.5\): \(f'(0.5) = 8(0.5)^3 - 8(0.5) = 1 - 4 = -3\) (negative)- In the interval \((1, \infty)\), pick \(x = 2\): \(f'(2) = 8(2)^3 - 8(2) = 64 - 16 = 48\) (positive)This shows changes from negative to positive at \(x = -1\) and from positive to negative at \(x = 0\), from negative to positive at \(x = 1\).

Step 5: Conclude with the results

The derivative \( f'(x) \) changes from negative to positive at \( x = -1 \) and \( x = 1 \), and from positive to negative at \( x = 0 \). Therefore, these are the values of \( c \) where the sign changes.

Key Concepts

Derivative

In calculus, the derivative of a function is a concept that measures how the function changes as its input changes. It is essentially the slope of the function at any given point. This is vital for understanding the behavior of functions, such as where a function is increasing or decreasing.

To find the derivative of a function, you use specific rules and techniques. For example, the derivative of a polynomial can be calculated using the power rule, which we will explore further. In our problem, given the function:

• \( f(x) = 2x^4 - 4x^2 + 3 \)

The derivative, \( f'(x) \), is calculated by applying the power rule to each term, resulting in:

• \( f'(x) = 8x^3 - 8x \)

This expression \( f'(x) \) helps us find the points where the function either increases or decreases, helping us locate the critical points, which are essential for understanding sign changes.

Sign Change

In calculus, sign changes in the derivative indicate where a function shifts from increasing to decreasing or vice versa. Understanding these changes can help you find the critical points that are essential in analyzing the function's behavior.

• When the derivative changes from positive to negative, it implies a local maximum at that point.
• Conversely, when the derivative changes from negative to positive, it signifies a local minimum.

For our exercise, having calculated the derivative \( f'(x) = 8x^3 - 8x \), we set it to zero to find where the sign might change:

• \( 8x^3 - 8x = 0 \)

Examining intervals around the critical points aids in determining these sign changes:

• Negative to positive at \( x = -1 \)
• Positive to negative at \( x = 0 \)
• Negative to positive at \( x = 1 \)

Through this process, you identify where the sign of the derivative changes, leading to insights about the shape and turning points of the function.

Power Rule

The power rule is a fundamental differentiation technique in calculus, used to find the derivative of functions of the form \( x^n \). It's one of the simplest and most important rules, allowing us to quickly determine derivatives.

The rule states:

• If \( n \) is a constant, then the derivative of \( x^n \) is \( nx^{n-1} \).

This can be applied term by term to each component of a polynomial.

For example, consider the function \( f(x) = 2x^4 - 4x^2 + 3 \):

• The derivative of \( 2x^4 \) is \( 8x^3 \) thanks to the power rule.
• Similarly, \(-4x^2 \) differentiates to \(-8x \).
• The constant 3 differentiates to 0, as constants disappear.

This results in the complete derivative \( f'(x) = 8x^3 - 8x \). Understanding how to use the power rule efficiently saves time and simplifies the process of finding derivatives, particularly for polynomial functions.

Factorization

Factorization is a mathematical process used to break down expressions into simpler components or products. It's crucial in solving equations, especially when trying to find roots.

In our exercise, after calculating the derivative \( f'(x) = 8x^3 - 8x \), we set it to zero to find the critical points:

• \( 8x^3 - 8x = 0 \)

The factorization process involves simplifying this equation:

• First, we factor out the common term \( 8x \):
• \( 8x(x^2 - 1) = 0 \)
• This yields components \( x = 0 \) and \( x^2 - 1 = 0 \).
• Further factor \( x^2 - 1 \) as \( (x - 1)(x + 1) = 0 \).
• Thus, giving solutions \( x = 1 \) and \( x = -1 \).

These critical points \( x = -1, 0, 1 \) highlight where the derivative's sign can shift, marking important spots of change in the function's behavior. Factorization not only aids in solving for these points but also in simplifying more complex expressions.