Problem 27 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=x^{4}-8 x^{2}+16$$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -2)\), \((-2, 0)\), and \((2, \infty)\), decreasing on \((0, 2)\). Local maxima at \(x = 0\) (16), minima at \(x = -2, 2\) (0).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. The derivative of the function, \( f(x) = x^4 - 8x^2 + 16 \), is found by differentiating term-by-term: \( f'(x) = 4x^3 - 16x \).

Step 2: Set the First Derivative to Zero

Next, we find where the derivative equals zero to identify critical points. Set \( f'(x) = 4x^3 - 16x = 0 \). Factor out \( 4x \): \( 4x(x^2 - 4) = 0 \). This gives us \( x = 0 \) or \( x^2 - 4 = 0 \). Solving \( x^2 - 4 = 0 \) gives \( x = 2 \) and \( x = -2 \). Thus, the critical points are \( x = -2, 0, 2 \).

Step 3: Identify Intervals using Critical Points

The critical points \( x = -2, 0, 2 \) divide the number line into the intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\).

Step 4: Determine Increasing/Decreasing Intervals

Test each interval by selecting a test point from each and plugging it into \( f'(x) \): - For \( x = -3 \) in \((-\infty, -2)\), \( f'(-3) = 108 \) (positive), so \( f(x) \) is increasing.- For \( x = -1 \) in \((-2, 0)\), \( f'(-1) = 12 \) (positive), so \( f(x) \) is increasing.- For \( x = 1 \) in \((0, 2)\), \( f'(1) = -12 \) (negative), so \( f(x) \) is decreasing.- For \( x = 3 \) in \((2, \infty)\), \( f'(3) = 72 \) (positive), so \( f(x) \) is increasing.

Step 5: Determine Local and Absolute Extrema

Evaluate the function at critical points:- \( f(-2) = (-2)^4 - 8(-2)^2 + 16 = 0 \) - \( f(0) = (0)^4 - 8(0)^2 + 16 = 16 \)- \( f(2) = (2)^4 - 8(2)^2 + 16 = 0 \)Since \( f(x) \) increases then decreases at each end as \( x \to \pm \infty \), there are no absolute extrema at infinity. The local maxima occur at \( x = 0 \) where \( f(0) = 16 \), and local minima occur at \( x = -2 \) and \( x = 2 \) where \( f = 0 \).

Key Concepts

Critical Points

A critical point of a function is where its derivative is either zero or undefined. These points are essential because they help us understand where a function might change direction. For the function in our exercise, we found the derivative to be \( f'(x) = 4x^3 - 16x \). We set this derivative to zero to find the critical points. This gives us the equation \( 4x(x^2 - 4) = 0 \).

• Setting \( 4x = 0 \), we get \( x = 0 \).
• Solving \( x^2 - 4 = 0 \) yields \( x = 2 \) and \( x = -2 \).

Thus, the critical points for our function are \( x = -2, 0, 2 \). These points are where the function's behavior might change from increasing to decreasing or vice versa.

Derivative

The derivative of a function provides a way to measure how it is changing at any given point. It tells us the slope or rate of change and is crucial for identifying where the function is increasing or decreasing. For the function \( f(x) = x^4 - 8x^2 + 16 \), the derivative is calculated as follows:

• The derivative of \( x^4 \) is \( 4x^3 \).
• The derivative of \( -8x^2 \) is \( -16x \).
• The derivative of a constant \( 16 \) is zero.

Putting it all together, the derivative of the function is \( f'(x) = 4x^3 - 16x \).
Understanding the derivative helps us determine where the function is flat, which is crucial for finding critical points.

Increasing and Decreasing Intervals

An increasing interval on a function is where the function values go up as \( x \) increases. Conversely, a decreasing interval is where the function values go down as \( x \) increases. To find these intervals, we look at the sign of the derivative.
We divide the number line using the critical points \( x = -2, 0, 2 \) into intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\). By choosing test points from each interval and substituting them into \( f'(x) \), we can determine:

• On \((-\infty, -2)\), \( f'(x) \) is positive, so the function is increasing.
• On \((-2, 0)\), \( f'(x) \) is positive, still increasing.
• On \((0, 2)\), \( f'(x) \) is negative, so the function is decreasing.
• On \((2, \infty)\), \( f'(x) \) is positive again, thus increasing.

Knowing which intervals show increasing or decreasing behavior helps us understand the overall shape of the graph.

Local Extrema

Local extrema refer to points on a function where it reaches a local maximum or minimum value. A local maximum is a point where the function is higher than all nearby points, while a local minimum is lower than any near points.
We found the critical points \( x = -2, 0, 2 \) through the derivative. By evaluating the function at these points, we identify whether they are local maxima or minima:

• At \( x = 0 \), \( f(0) = 16 \) is a local maximum because the function decreases immediately after.
• At \( x = -2 \), \( f(-2) = 0 \) is a local minimum since the function was increasing before and decreases after.
• At \( x = 2 \), \( f(2) = 0 \) is another local minimum for similar reasoning as \( x = -2 \).

Understanding where these local extrema occur assists in sketching the function and predicting its behavior. There are no absolute maximum or minimum values as \( x \to \pm\infty \), meaning the function's values do not reach extreme values beyond these local ones.