Given the function y = x^4- 8x² + 16. on which intervals is the function increasing? a. empty set b. (-infin, infin) c (-infin, -2) and (0,2) d. (-2,0) and (2, infin)

Step-by-step explanation:

Recall that we use the first derivative to discover where a function is increasing or decreasing; it's increasing where the first derivative is + and decreasing where the first derivative is -.

The derivative of y = x^4 - 8x^2 + 16 is  dy/dx = 4x^3 - 16x, or 4x(x^2 - 4).

This can be factored further:  4x(x - 2)(x + 2).

Set this equal to zero and solve for the three critical values:

{-2, 0, 2}.

Set up a total of four intervals:  (-∞, -2), (-2, 0), (0, 2), (2, ∞).

Now choose a test point within each interval:  {-3, -1, 1, 3}.

Evaluate the first derivative, dy/dx, at each of these four test points.  Rule:  if the first derivative is +, we know the function is increasing on that interval; if -, the function is decreasing.

At x = -3, dy/dx = 4x(x - 2)(x + 2) becomes (-)(-)(-), which is -, so we know that the function is decreasing on interval (∞, -2).

At x = -1, dy/dx is (-)(-)(+), which is +, so we know that the function is increasing on (-2, 0).

At x = 1, dy/dx is (+)(-)(+), which is -, so we know that the function is decreasing on (0, 2).

Finally:  at x = 3, dy/dx is (+)(+)(+), so we know that the function is increasing on (2, ∞ ).

answer: 12

step-by-step explanation:

because yes