Problem 17 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ f(x)=x^{4}-8 x^{2}+16 $$

Step-by-Step Solution

Answer

Increasing: (-∞, -2) ∪ (2, ∞). Decreasing: (-2, 0) ∪ (0, 2). Local max at \(x=-2\) is 16. Local and absolute min at \(x=2\) is 0.

Step 1: Find the derivative of the function

To analyze the intervals of increase and decrease, first find the derivative of the function: \[ f(x) = x^4 - 8x^2 + 16 \] The derivative \( f'(x) \) is \[ f'(x) = \frac{d}{dx}(x^4 - 8x^2 + 16) = 4x^3 - 16x \]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Since polynomials are defined everywhere, set the derivative equal to zero to find critical points:\[ 4x^3 - 16x = 0 \] Factor out the common term:\[ 4x(x^2 - 4) = 0 \] This gives critical points:\[ x = 0, \pm2 \]

Step 3: Determine intervals of increase and decrease

Use the critical points to test intervals on the number line: (-∞, -2), (-2, 0), (0, 2), and (2, ∞). Test a point from each interval in the derivative:1. Choose \(x = -3\) from (-∞, -2): \(f'(-3) = 4(-3)^3 - 16(-3) > 0\), the function is increasing.2. Choose \(x = -1\) from (-2, 0): \(f'(-1) = 4(-1)^3 - 16(-1) < 0\), the function is decreasing.3. Choose \(x = 1\) from (0, 2): \(f'(1) = 4(1)^3 - 16(1) < 0\), the function is decreasing.4. Choose \(x = 3\) from (2, ∞): \(f'(3) = 4(3)^3 - 16(3) > 0\), the function is increasing.

Step 4: Identify the local extreme values

By examining sign changes in \(f'(x)\):- At \(x = -2\), \(f'(x)\) changes from positive to negative, indicating a local maximum.- At \(x = 0\), \(f'(x)\) does not change sign, hence no extremum.- At \(x = 2\), \(f'(x)\) changes from negative to positive, indicating a local minimum.Evaluate the function at these points to find values:\(f(-2) = (-2)^4 - 8(-2)^2 + 16 = 16\), thus a local maximum.\(f(2) = (2)^4 - 8(2)^2 + 16 = 0\), thus a local minimum.

Step 5: Determine absolute extremes

Consider both endpoints and infinity: - As \(x \to \pm \infty\), \(f(x) \to \infty\) since the leading term \(x^4\) dominates. - Therefore, the absolute minimum occurs at \(x = 2\) with the value \(f(2) = 0\). - There is no absolute maximum as the function approaches infinity both ways.

Step 6: Graph the function

You can verify the above findings using a graphing calculator or computer grapher by plotting \(f(x) = x^4 - 8x^2 + 16\). The graph should display a high point at \(x = -2\) (local maximum), a low point at \(x = 2\) (local and absolute minimum), and should reveal that the function increases as \(x\) approaches \(\pm∞\).

Key Concepts

Derivatives

To understand changes in a function, we first look at its derivative. A derivative represents how a function's output, or value, changes concerning changes in its input, or variable. For the function \( f(x) = x^4 - 8x^2 + 16 \), we find its derivative by applying standard differentiation rules. The derivative is \( f'(x) = 4x^3 - 16x \).
This is obtained by differentiating each term separately. The derivative gives us insights into how the function behaves—whether it's increasing, decreasing, or remaining constant at different points along the curve.

Critical Points

Critical points are crucial when analyzing a function. These points occur where the derivative is either zero or undefined. Since polynomials are defined everywhere, we find critical points by setting the derivative \( f'(x) = 4x^3 - 16x \) equal to zero.

• Factorizing gives us \( 4x(x^2 - 4) = 0 \), leading to critical points at \( x = 0, \pm2 \).

These are the points where potential changes in the function's behavior occur, like swapping from increasing to decreasing or vice versa.

Intervals of Increase and Decrease

The derivative tells us how the function is changing. By analyzing the sign of \( f'(x) \) across different intervals, we understand when the function is increasing or decreasing. On the number line, these intervals are (-∞, -2), (-2, 0), (0, 2), and (2, ∞):

• For \( x eq -2 \), where \( f'(-3) > 0 \), the function is increasing.
• For \( -2 < x < 0 \), where \( f'(-1) < 0 \), the function is decreasing.
• For \( 0 < x < 2 \), where \( f'(1) < 0 \), the function keeps decreasing.
• For \( x > 2 \), where \( f'(3) > 0 \), the function is increasing again.

Local Extrema

Local extrema refer to points on a function where it reaches a local maximum or minimum value. They are identified at the critical points by checking where the sign of the derivative changes:

• At \( x = -2 \), the derivative changes from positive to negative, indicating a local maximum.
• At \( x = 0 \), the sign of the derivative remains unchanged, showing no extremum.
• At \( x = 2 \), the derivative changes from negative to positive, indicating a local minimum.

By evaluating \( f(x) \) at these points, such as \( f(-2) = 16 \) and \( f(2) = 0 \), the actual values of the local maximum and minimum can be confirmed.

Absolute Extrema

Absolute extrema identify the highest or lowest points a function reaches on its entire domain. In this case:

• The function \( f(x) \) behaves differently as \( x \to \pm \infty \), suggesting no absolute maximum since it heads towards infinity.
• An absolute minimum is observed at \( x = 2 \) with the value \( f(2) = 0 \), confirmed by its minimum local value.

These conclusions can be visually supported by plotting the graph, which shows trends depicted by the derivative analysis, verifying absolute behavior.