Find:
(a) the interval(s) on which f is increasing,
(b) the interval(s) on which f is decreasing,
(c) the open interval(s) on which f is concave up,
(d) the open interval(s) on which f is concave down, and
(e) the X-coordinate(s) of any/all inflection point(s).
f(x)= (x^4) - (4802 x^2) + 9604
To find the intervals on which f is increasing or decreasing, we can find the derivative of f and analyze its sign.
First, let's find the derivative of f(x):
f'(x) = 4x^3 - 9604x
To determine the intervals where f is increasing or decreasing, we need to analyze the sign of f'(x).
(a) To find the intervals where f is increasing, we need to find where f'(x) > 0.
Setting f'(x) > 0:
4x^3 - 9604x > 0
Factoring out 4x:
4x(x^2 - 2401) > 0
Now, we have two factors to consider separately:
1. 4x > 0
This implies x > 0.
2. (x^2 - 2401) > 0
Solving this inequality, we find:
x > 49 or x < -49
Therefore, the intervals where f is increasing are (0, 49) and (-∞, -49).
(b) To find the intervals where f is decreasing, we need to find where f'(x) < 0.
Setting f'(x) < 0:
4x^3 - 9604x < 0
Factoring out 4x:
4x(x^2 - 2401) < 0
Now, we have two factors to consider separately:
1. 4x < 0
This implies x < 0.
2. (x^2 - 2401) < 0
Solving this inequality, we find:
-49 < x < 49
Therefore, the interval where f is decreasing is (-49, 0).
To find the open intervals where f is concave up or concave down, we need to find the second derivative of f(x):
f''(x) = 12x^2 - 9604
Now, let's analyze the sign of f''(x) to determine the intervals of concavity.
(c) To find the open intervals where f is concave up, we need to find where f''(x) > 0.
Setting f''(x) > 0:
12x^2 - 9604 > 0
Simplifying:
3x^2 - 2401 > 0
Factoring:
(3x + 49)(x - 49) > 0
To make this inequality true, we need:
1. 3x + 49 > 0
This implies x > -49/3.
2. x - 49 > 0
This implies x > 49.
Therefore, the open interval where f is concave up is (49, ∞).
(d) To find the open intervals where f is concave down, we need to find where f''(x) < 0.
Setting f''(x) < 0:
12x^2 - 9604 < 0
Simplifying:
3x^2 - 2401 < 0
Factoring:
(3x + 49)(x - 49) < 0
To make this inequality true, we need:
1. 3x + 49 < 0
This implies x < -49/3.
2. x - 49 < 0
This implies x < 49.
Therefore, the open interval where f is concave down is (-∞, -49/3) U (-49, 49).
(e) To find the x-coordinate(s) of any inflection point(s), we need to find the x-values where the concavity changes. In this case, the concavity changes from concave down to concave up, or vice versa, at the x-values where f''(x) = 0.
Setting f''(x) = 0:
12x^2 - 9604 = 0
Simplifying:
3x^2 - 2401 = 0
Factoring:
(3x + 49)(x - 49) = 0
Solving for x, we find two solutions:
1. 3x + 49 = 0
This implies x = -49/3
2. x - 49 = 0
This implies x = 49
Therefore, the x-coordinates of the inflection points are -49/3 and 49.
To find the information about the intervals on which f is increasing or decreasing, and the intervals on which f is concave up or concave down, we need to determine the first and second derivatives of the function f(x).
(a) To find the intervals on which f is increasing:
We need to find the x-values where the first derivative is positive (greater than zero).
First, find the first derivative of f(x):
f'(x) = 4x^3 - 9604x
Now set f'(x) > 0 and solve for x:
4x^3 - 9604x > 0
x(4x^2 - 2401) > 0
The critical points of the first derivative occur when x = 0, x = sqrt(2401), and x = -sqrt(2401).
Create a sign chart and evaluate the sign of f'(x) in each interval:
Interval (-∞, -√(2401)) --> f'(x) is negative
Interval (-√(2401), 0) --> f'(x) is positive
Interval (0, √(2401)) --> f'(x) is negative
Interval (√(2401), ∞) --> f'(x) is positive
Therefore, f(x) is increasing on the intervals (-√(2401), 0) and (√(2401), ∞).
(b) To find the intervals on which f is decreasing:
We need to find the x-values where the first derivative is negative (less than zero).
Using the sign chart obtained from step (a), we can observe that f(x) is decreasing on the intervals (-∞, -√(2401)] and [0, √(2401)].
(c) To find the open intervals on which f is concave up:
We need to find the x-values where the second derivative is positive.
Find the second derivative of f(x):
f''(x) = 12x^2 - 9604
Set f''(x) > 0 and solve for x:
12x^2 - 9604 > 0
12x^2 > 9604
x^2 > 800.33
x > √(800.33) or x < -√(800.33)
Since we're looking for open intervals, we conclude that f(x) is concave up on the interval (-∞, -√(800.33)) U (√(800.33), ∞).
(d) To find the open intervals on which f is concave down:
We need to find the x-values where the second derivative is negative.
Using the information from step (c), we can observe that f(x) is concave down on the interval -√(800.33), √(800.33).
(e) To find the x-coordinate(s) of any inflection point(s):
Inflection points occur when the second derivative changes sign. Using the information from steps (c) and (d), we can conclude that f doesn't have an inflection point since f''(x) does not change sign between the intervals (-√(800.33), √(800.33)).
In summary:
(a) f(x) is increasing on the intervals (-√(2401), 0) and (√(2401), ∞).
(b) f(x) is decreasing on the intervals (-∞, -√(2401)] and [0, √(2401)].
(c) f(x) is concave up on the interval (-∞, -√(800.33)) U (√(800.33), ∞).
(d) f(x) is concave down on the interval -√(800.33), √(800.33).
(e) There are no inflection points.