Answer:
True
False
False
True
Step-by-step explanation: The y-intercept of the graph of f(x) is 3.
To find the y-intercept, we set x = 0 and evaluate f(0):
f(0) = 2(0)^3 + 3 = 3
So, the statement is true.
The graph of f(x) is symmetric about the y-axis.
To check for symmetry about the y-axis, we substitute -x for x in the equation for f(x):
f(-x) = 2(-x)^3 + 3 = -2x^3 + 3
If the graph of f(x) is symmetric about the y-axis, then f(-x) = f(x) for all x. However, in this case, we have:
f(-x) = -2x^3 + 3 ≠ f(x)
So, the statement is false.
The function f(x) has a local maximum at x = 0.
To find local maximum or minimum, we need to find the critical points by setting f'(x) = 0.
f(x) = 2x^3 + 3
f'(x) = 6x^2
Setting f'(x) = 0, we get:
6x^2 = 0
x = 0
So, x = 0 is a critical point. To determine whether it is a local maximum or minimum, we need to examine the sign of f''(x):
f''(x) = 12x
At x = 0, f''(0) = 0, which means that the second derivative test is inconclusive. Therefore, we cannot conclude whether x = 0 is a local maximum or minimum.
So, the statement is false.
The function f(x) is increasing on the interval (-∞, 0) and decreasing on the interval (0, ∞).
To determine the intervals of increase and decrease, we need to examine the sign of f'(x):
f'(x) = 6x^2
f'(x) is positive on the interval (0, ∞) and negative on the interval (-∞, 0). Therefore, f(x) is increasing on the interval (0, ∞) and decreasing on the interval (-∞, 0).
So, the statement is true.