To identify the intervals where the given function is increasing, decreasing, concave up, and concave down, as well as find the maximum and minimum values, we can follow these steps:
Step 1: Determine the first derivative to find intervals of increasing and decreasing.
a) Find the derivative of y = x^5 - 15x^3 + 10:
y' = 5x^4 - 45x^2
b) Set the derivative equal to zero and solve for x:
5x^4 - 45x^2 = 0
5x^2(x^2 - 9) = 0
x = 0 (This is not in the interval provided, so it will be excluded.)
x = ±3
The critical points are x = -3 and x = 3.
Step 2: Create a number line and choose test points to determine the intervals of increasing and decreasing.
Test point in the interval (1, 5):
If we test a value less than 3 (e.g., x = 2), we find that y' > 0, indicating increasing behavior.
If we test a value greater than 3 (e.g., x = 4), we find that y' > 0, indicating increasing behavior.
Therefore, the function is increasing on the interval (1, 5).
Step 3: Determine the second derivative to find intervals of concavity.
a) Find the second derivative of y':
y'' = 20x^3 - 90x
b) Set the second derivative equal to zero and solve for x:
20x^3 - 90x = 0
10x(x^2 - 9) = 0
x = 0, x = -3, and x = 3
The critical points are x = -3 (previously identified) and x = 0.
Step 4: Create a number line and choose test points to determine the concavity intervals.
Test points in the intervals (-∞, -3), (-3, 0), and (0, ∞):
For x < -3 (e.g., x = -4), we find that y'' < 0, indicating concave down behavior.
For -3 < x < 0 (e.g., x = -1), we find that y'' > 0, indicating concave up behavior.
For x > 0 (e.g., x = 1), we find that y'' > 0, indicating concave up behavior.
Therefore, the function is concave down on the interval (-∞, -3), concave up on the interval (-3, 0), and concave up on the interval (0, ∞). Since the critical points -3 and 3 are within the interval (1, 5), these are also the inflection points.
Step 5: Find the maximum and minimum values of y within the restricted interval (1, 5).
Evaluate the function at the endpoints and critical points:
For x = 1:
y = 1^5 - 15(1^3) + 10 = -4
For x = 5:
y = 5^5 - 15(5^3) + 10 = 4640
For x = -3:
y = (-3)^5 - 15(-3)^3 + 10 = -328
For x = 3:
y = 3^5 - 15(3)^3 + 10 = 388
The maximum value within the interval (1, 5) is 4640, which occurs at x = 5. The minimum value is -328, which occurs at x = -3.
By following these steps and calculations, you should be able to identify the intervals of increasing, decreasing, concave up, and concave down, plot the critical points, and find the maximum and minimum values for the given function.