Answer:
(a) Increasing in the interval (-623, 2) and (623, ∞)
f is decreasing in the interval (∞, -623) and (2, - 623).
(b) Local maximum is f = 2 .
2 local minima f = -623.
Step-by-step explanation:
f(x) = x^4 - 50x^2 + 2
Finding the derivative:
f'(x) = 4x^3 - 100x
This = 0 at a turning point:
4x(x^2 - 25) = 0
x = 0, x + -5, 5.
Find the second derivative:
f"(x) = 12x^2 - 100
When x = 0 f(x) = 2
When x = 0, f"(x) is negative so the point (0, 2) is a maximum .
When x = -5, f"(x) is positive . Also positive when x = 5.
So the 2 local minimums are (-5, (-5)^4 - 50(-5)^2 + 2) = (-5, -623) and
(5, -623).
