Consider the equation below. (if an answer does not exist, enter dne.) f(x) = x4 − 50x2 + 2 (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the interval on which f is decreasing. (enter your answer using interval notation.) (b) find the local minimum and maximum values of f. local minimum value local maximum value (c) find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) find the interval on which f is concave up. (enter your answer using interval notation.) find the interval on which f is concave down.(enter your answer using interval notation.)

(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).

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