Consider the function below. (Round the answers to three decimal places. If you need to use - or , enter -INFINITY or INFINITY.)
f(x) = 4 + 4x2 - x4
(a) Find the intervals of increase. (Enter the interval that contains smaller numbers first.)
Find the intervals of decrease. (Enter the interval that contains smaller numbers first.)
(b) Find the local minimum value.
Find the local maximum values.
(smaller x value)
(larger x value)
(c) Find the inflection points.
(smaller x value)
(larger x value)
Find the interval the function is concave up.
Find the intervals the function is concave down. (Enter the interval that contains smaller numbers first.)
((d) Use this information to sketch the graph of the function.
To find the intervals of increase and decrease, we need to determine where the derivative of the function is positive or negative.
(a) Intervals of Increase and Decrease:
1. Take the derivative of the function f(x) with respect to x:
f'(x) = 8x - 4x^3
2. Set the derivative equal to zero and solve for x to find critical points:
8x - 4x^3 = 0
Factor out 4x: 4x(x^2 - 1) = 0
This gives two critical points: x = 0 and x = ±1.
3. Create a sign chart by picking test points within each interval separated by the critical points.
Test x = -2: f'(-2) = (-16) - 4(-8) = -16 + 32 = 16 (positive)
Test x = -0.5: f'(-0.5) = (-4) - 4(0.25) = -4 - 1 = -5 (negative)
Test x = 0.5: f'(0.5) = (4) - 4(0.25) = 4 - 1 = 3 (positive)
Test x = 2: f'(2) = (16) - 4(8) = 16 - 32 = -16 (negative)
Based on the sign chart, the intervals of increase are: (-∞, -1) and (1, ∞)
The intervals of decrease are: (-1, 1)
(b) Local Minimum Value:
To find the local minimum value, we need to find the y-coordinate corresponding to the local minimum.
1. Evaluate the function f(x) at the critical points:
f(0) = 4 + 4(0)^2 - (0)^4 = 4 + 0 - 0 = 4
f(1) = 4 + 4(1)^2 - (1)^4 = 4 + 4 - 1 = 7
f(-1) = 4 + 4(-1)^2 - (-1)^4 = 4 + 4 - 1 = 7
The local minimum value is 4.
(c) Inflection Points:
To find the inflection points, we need to determine where the concavity of the function changes.
1. Take the second derivative of the function f(x) with respect to x:
f''(x) = 8 - 12x^2
2. Set the second derivative equal to zero and solve for x to find possible inflection points:
8 - 12x^2 = 0
Divide by 4: 2 - 3x^2 = 0
3x^2 = 2
x^2 = 2/3
x = ±√(2/3)
These give two possible inflection points: x = -√(2/3) and x = √(2/3).
3. Create a sign chart by picking test points within each interval separated by the possible inflection points.
Test x = -2: f''(-2) = 8 - 12(4) = 8 - 48 = -40 (negative)
Test x = -1: f''(-1) = 8 - 12(1) = 8 - 12 = -4 (negative)
Test x = 0: f''(0) = 8 - 12(0) = 8 (positive)
Test x = 1: f''(1) = 8 - 12(1) = 8 - 12 = -4 (negative)
Test x = 2: f''(2) = 8 - 12(4) = 8 - 48 = -40 (negative)
Based on the sign chart, the intervals where the function is concave up are (-∞, -√(2/3)) and (√(2/3), ∞).
The interval where the function is concave down is (-√(2/3), √(2/3)).
(d) Sketching the Graph:
Based on the information gathered, we can sketch the graph of the function.
- The graph increases on the intervals (-∞, -1) and (1, ∞).
- The graph decreases on the interval (-1, 1).
- The graph has a local minimum value at (0, 4).
- The graph has an inflection point at x = -√(2/3).
- The graph has another inflection point at x = √(2/3).
- The graph is concave up on the intervals (-∞, -√(2/3)) and (√(2/3), ∞).
- The graph is concave down on the interval (-√(2/3), √(2/3)).
Note: It is recommended to use a graphing tool or software to accurately visualize the graph.
To find the intervals of increase and decrease, we need to find the derivative of the function f(x) and determine where it is positive or negative.
(a) Taking the derivative of f(x), we get:
f'(x) = 8x - 4x^3
Setting f'(x) = 0, we can solve for critical points:
8x - 4x^3 = 0
4x(2 - x^2) = 0
From this equation, we find critical points at x = 0, x = -√2, and x = √2. These points divide the number line into intervals:
Interval of increase: (-∞, -√2) U (√2, ∞)
Interval of decrease: (-√2, 0) U (0, √2)
(b) To find the local minimum value, we need to identify where the function changes from decreasing to increasing. The first derivative test can be used to determine this.
At x = -√2, f'(-√2) = 8(-√2) - 4(-√2)^3
= -8√2 + 4√8
= -8√2 + 8√2
= 0
At x = √2, f'(√2) = 8(√2) - 4(√2)^3
= 8√2 - 4√8
= 8√2 - 8√2
= 0
Since f'(x) changes from negative to positive at x = -√2 and from positive to negative at x = √2, these are the potential local minimum points.
To determine the minimum values, we evaluate f(x) at these points:
f(-√2) = 4 + 4(-√2)^2 - (-√2)^4
= 4 + 4(2) - 2
= 12
f(√2) = 4 + 4(√2)^2 - (√2)^4
= 4 + 4(2) - 2
= 12
Therefore, the local minimum value is 12.
(c) To find the inflection points, we need to find the second derivative of f(x) and determine where it changes sign.
Taking the second derivative:
f''(x) = 8 - 12x^2
Setting f''(x) = 0, we can solve for critical points:
8 - 12x^2 = 0
12x^2 = 8
x^2 = 2/3
x = ±√(2/3)
So, the inflection points are at x = -√(2/3) and x = √(2/3).
To determine the concavity, we look at the intervals between these critical points:
Interval of concave up: (-∞, -√(2/3)) U (√(2/3), ∞)
Interval of concave down: (-√(2/3), √(2/3))
(d) Based on the information we have gathered, we can now sketch the graph of the function f(x). The graph will have a local minimum at x = -√2 and x = √2, and inflection points at x = -√(2/3) and x = √(2/3). The concavity changes at these inflection points.
However, we still need to determine the behavior of the function as x approaches ±∞. By observing the highest-degree term in f(x), which is -x^4, we can see that as x approaches ±∞, f(x) approaches -∞.
Putting all of this information together, we can sketch the graph of the function f(x) with the following characteristics:
- Local minimum: (x, y) = (-√2, 12) and (x, y) = (√2, 12)
- Inflection points: (x, y) = (-√(2/3), f(-√(2/3))) and (x, y) = (√(2/3), f(√(2/3)))
- Interval of concave up: (-∞, -√(2/3)) U (√(2/3), ∞)
- Interval of concave down: (-√(2/3), √(2/3))
- Asymptote: As x approaches ±∞, f(x) approaches -∞.