How to analytically find the intervals on which the function y = 2x^4 - 4x^2 + 1 is

1)increasing
2)decreasing
3)concave up
4)concave down

Also find any local extreme values and inflection points.

Thanks.

(1) The function is increasing when the first derivative dy/dx = 8x^3 - 8x is positive
(2) The function is decreasing there the same derivative is negative.
(3) The function is concave up where the second derivative, 24 x^2 - 8, is positive.
(4) The function is concave down where the second derivative is negative.
Extreme values occur whereever dy/dx = 0
Inflection points are where the second deritivative is zero.

The rest is algebra.

To find the intervals on which the function is increasing or decreasing, we need to find the derivative of the function, y = 2x^4 - 4x^2 + 1, with respect to x:

dy/dx = d/dx (2x^4 - 4x^2 + 1)
= 8x^3 - 8x

To find where the function is increasing, we set the derivative to be greater than zero:

8x^3 - 8x > 0

We can factor out 8x from the expression:

8x(x^2 - 1) > 0

Now, we have two factors, 8x and (x^2 - 1). To satisfy the inequality, either both factors must be greater than zero or both factors must be less than zero.

Considering the first factor, 8x > 0, we see that x > 0 for it to be greater than zero.

For the second factor, (x^2 - 1) > 0, we can factor it further:

(x - 1)(x + 1) > 0

This means that either both (x - 1) and (x + 1) are greater than zero or both are less than zero. Analyzing the signs of both factors, we find that x > 1 or x < -1.

Putting all the conditions together, we find that the function is increasing for x > 1 and decreasing for x < -1.

To determine where the function is concave up or concave down, we need to find the second derivative. Taking the derivative of the first derivative, we get:

d^2y/dx^2 = d/dx (8x^3 - 8x)
= 24x^2 - 8

For the function to be concave up, the second derivative must be greater than zero:

24x^2 - 8 > 0

Simplifying the inequality:

24x^2 > 8
x^2 > 8/24
x^2 > 1/3

Taking the square root to isolate x:

x > √(1/3) or x < -√(1/3)

This means that the function is concave up for x > √(1/3) or x < -√(1/3).

For the function to be concave down, the second derivative must be less than zero:

24x^2 - 8 < 0

Simplifying the inequality:

24x^2 < 8
x^2 < 8/24
x^2 < 1/3

This indicates that the function is concave down for -√(1/3) < x < √(1/3).

To find the local extreme values, we set the first derivative equal to zero:

8x^3 - 8x = 0

Factoring out 8x:

8x(x^2 - 1) = 0

Setting each factor to zero:

8x = 0 or x^2 - 1 = 0

For the first factor, 8x = 0, x = 0.

For the second factor, x^2 - 1 = 0, we factor it:

(x - 1)(x + 1) = 0

Setting each factor to zero:

x - 1 = 0 or x + 1 = 0

x = 1 or x = -1

Therefore, the function has local extreme values at x = 0, x = 1, and x = -1.

To find the inflection points, we set the second derivative equal to zero:

24x^2 - 8 = 0

Simplifying the equation:

24x^2 = 8
x^2 = 8/24
x^2 = 1/3

Taking the square root to isolate x:

x = ±√(1/3)

Hence, the function has inflection points at x = √(1/3) and x = -√(1/3).

In summary:
1) The function is increasing for x > 1 and decreasing for x < -1.
2) The function is concave up for x > √(1/3) or x < -√(1/3).
3) The function is concave down for -√(1/3) < x < √(1/3).
4) The local extreme values occur at x = 0, x = 1, and x = -1.
5) The inflection points occur at x = √(1/3) and x = -√(1/3).