Problem 27 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one. \(f(x)=2+2 x^{2}-x^{4}\)

Step-by-Step Solution

Answer

Increase: (-1, 0) and (0, 1); Decrease: (-∞, -1) and (1, ∞). Local max at x=1, min at x=-1. Inflection at x=±√(1/3).

Step 1: Find the first derivative

To determine the intervals of increase or decrease, we first need the first derivative of the function. For the given function \( f(x) = 2 + 2x^2 - x^4 \), differentiate with respect to \( x \). This gives:\[ f'(x) = 4x - 4x^3 \]

Step 2: Solve the critical points

Set the first derivative equal to zero to find the critical points: \[ 4x - 4x^3 = 0 \]Factor the equation:\[ 4x(1 - x^2) = 0 \]Solving for \( x \), we get: \( x = 0 \) or \( x = \pm 1 \).

Step 3: Test intervals around critical points for increase or decrease

Test the sign of \( f'(x) \) in the intervals determined by the critical points (-∞, -1), (-1, 0), (0, 1), and (1, ∞):- For \( x < -1 \), choose \( x = -2 \): \( f'(-2) = 4(-2) - 4(-2)^3 = -32 \), function is decreasing.- For \( -1 < x < 0 \), choose \( x = -0.5 \): \( f'(-0.5) = 4(-0.5) - 4(-0.5)^3 = 1.5 \), function is increasing.- For \( 0 < x < 1 \), choose \( x = 0.5 \): \( f'(0.5) = 4(0.5) - 4(0.5)^3 = 1.5 \), function is increasing.- For \( x > 1 \), choose \( x = 2 \): \( f'(2) = 4(2) - 4(2)^3 = -32 \), function is decreasing.

Step 4: Identify local maximum and minimum values

Using the sign changes from step 3:- At \( x = -1 \), \( f'(x) \) changes from negative to positive, thus \( x = -1 \) is a local minimum.- At \( x = 0 \), \( f'(x) \) does not change sign.- At \( x = 1 \), \( f'(x) \) changes from positive to negative, thus \( x = 1 \) is a local maximum.

Step 5: Find the second derivative for concavity

Find the second derivative of \( f(x) \) to analyze concavity:\[ f''(x) = 4 - 12x^2 \]

Step 6: Solve the inflection points

Set the second derivative equal to zero to find potential inflection points:\[ 4 - 12x^2 = 0 \]Solving for \( x \), we get:\[ x = \pm \sqrt{\frac{1}{3}} \].

Step 7: Test intervals around points for concavity

Determine concavity between and beyond inflection points \(-\sqrt{\frac{1}{3}}\), \(\sqrt{\frac{1}{3}}\):- For \( x < -\sqrt{\frac{1}{3}} \), test \( x = -1 \): \( f''(-1) = -8 \), concave down.- For \( -\sqrt{\frac{1}{3}} < x < \sqrt{\frac{1}{3}} \), test \( x = 0 \): \( f''(0) = 4 \), concave up.- For \( x > \sqrt{\frac{1}{3}} \), test \( x = 1 \): \( f''(1) = -8 \), concave down.

Step 8: Sketch the graph

Using the information from the previous steps, sketch the graph:- Increase: \((-1, 0) \text{ and } (0, 1)\);- Decrease: \((- ext{∞}, -1) \text{ and } (1, ext{∞})\);- Local max at \(x = 1\), local min at \(x = -1\);- Concave up: \((-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}})\);- Concave down: \((- ext{∞}, -\sqrt{\frac{1}{3}}) \text{ and } (\sqrt{\frac{1}{3}}, \text{∞})\).

Key Concepts

Critical Points

Critical points are values of \( x \) where the first derivative of a function equals zero or is undefined. These points help us identify where a function could potentially reach a local maximum or minimum. In the exercise, we find the critical points by setting the first derivative, \( f'(x) = 4x - 4x^3 \), equal to zero.

• We factor to get \( 4x(1 - x^2) = 0 \).
• This gives us the critical points \( x = 0 \) and \( x = \pm 1 \).

These critical points are indicators of where the function changes direction from increasing to decreasing, or vice versa. They are essential for sketching graphs and understanding the behavior of the function in different intervals.

Intervals of Increase and Decrease

To determine intervals where the function increases or decreases, evaluate the sign of the first derivative, \( f'(x) \), around the critical points. Divide the number line into intervals using the critical points found: \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, \infty)\). Test points from within each interval in the first derivative:

• For \(x < -1\), \(f'(-2) = -32\), indicating the function is decreasing.
• For \(-1 < x < 0\), \(f'(-0.5) = 1.5\), indicating the function is increasing.
• For \(0 < x < 1\), \(f'(0.5) = 1.5\), indicating the function is increasing.
• For \(x > 1\), \(f'(2) = -32\), indicating the function is decreasing.

Understanding these intervals allows you to sense when the function is climbing up or sliding down, essential for graphing and interpreting the function's behavior. Changes in interval signs also signal possible turning points.

Concavity and Inflection Points

Concavity refers to whether a function curves upwards or downwards. Use the second derivative, \( f''(x) \), to analyze this property. For the function \( f(x) = 2 + 2x^2 - x^4 \), the second derivative is \( f''(x) = 4 - 12x^2 \). To discover inflection points, set \( f''(x) \) to zero:

• \( 4 - 12x^2 = 0 \) gives \( x = \pm \sqrt{\frac{1}{3}} \).

These points are where the curvature changes from upwards to downwards or vice versa. Between intervals, test values:

• For \( x < -\sqrt{\frac{1}{3}} \), \(f''(-1) = -8\), so the curve is concave down.
• For \( -\sqrt{\frac{1}{3}} < x < \sqrt{\frac{1}{3}} \), \(f''(0) = 4\), so the curve is concave up.
• For \( x > \sqrt{\frac{1}{3}} \), \(f''(1) = -8\), so the curve is concave down.

These inflection points and the concavity analysis are keys to understanding shifts in a graph's bending, helping sketch an accurate representation of the function. They determine where the graph changes from bending one way to the other.