Final answer:
To find the values of x where f(x) = 3x^5 - 5x^3 + 15 has a relative maximum, we need to find the critical points of the function. By taking the derivative and evaluating the second derivative, we find that the function has relative maxima at x = -∞ to -1 and x = 0 to ∞.
Step-by-step explanation:
The function f(x) = 3x^5 - 5x^3 + 15 has a relative maximum at the values of x where the derivative of the function changes from positive to negative.
To find these values, we need to find the critical points of the function. Taking the derivative of f(x) and setting it equal to zero, we get:
f'(x) = 15x^4 - 15x^2 = 0
Factoring out 15x^2 from the equation, we have:
15x^2(x^2 - 1) = 0
Solving for x, we find two critical points: x = 0 and x = ±1.
These critical points divide the x-axis into three intervals: (-∞, -1), (-1, 0), and (0, ∞).
To determine whether these critical points correspond to a relative maximum or minimum, we can evaluate the second derivative of f(x) at each critical point:
f''(x) = 60x^3 - 30x
For x = -∞ to -1, f''(x) is negative, indicating a concave downward and a relative maximum. For x = -1 to 0, f''(x) is positive, indicating a concave upward and a relative minimum.
Lastly, for x = 0 to ∞, f''(x) is negative again, indicating a concave downward and a relative maximum.
Therefore, the values of x where f(x) = 3x^5 - 5x^3 + 15 has a relative maximum are x = -∞ to -1 and x = 0 to ∞.