For the function y=(x^4)-(2x^2)+1 Identify all relative extrema. Identify any points of inflection. Identify the absolute extrema on the interval[-2,3].
To find the relative extrema of a function, we first need to find the critical points of the function by taking its derivative and setting it equal to zero.
1. Find the derivative of the function:
The derivative of y = x^4 - 2x^2 + 1 is y' = 4x^3 - 4x.
2. Set the derivative equal to zero and solve for x:
4x^3 - 4x = 0.
Factor out 4x:
4x(x^2 - 1) = 0.
Set each factor equal to zero:
- 4x = 0 → x = 0,
x^2 - 1 = 0 → x^2 = 1 → x = ±1.
The critical points of the function are when x = -1, 0, and 1.
3. Determine the nature of each critical point:
We can use the second derivative test to determine the nature of each critical point.
The second derivative of y = x^4 - 2x^2 + 1 is y'' = 12x^2 - 4.
- For x = -1:
Substituting x = -1 into the second derivative, we get y'' = 12(-1)^2 - 4 = 8. Since the second derivative is positive, the point x = -1 is a relative minimum.
- For x = 0:
Substituting x = 0 into the second derivative, we get y'' = 12(0)^2 - 4 = -4. Since the second derivative is negative, the point x = 0 is a relative maximum.
- For x = 1:
Substituting x = 1 into the second derivative, we get y'' = 12(1)^2 - 4 = 8. Since the second derivative is positive, the point x = 1 is a relative minimum.
Therefore, the relative extrema of the function y = x^4 - 2x^2 + 1 are:
Relative maximum at (0, 1).
Relative minimums at (-1, 0) and (1, 0).
To find the points of inflection, we need to find where the concavity of the function changes. This occurs when the second derivative is equal to zero or does not exist.
4. Find the points of inflection:
Set the second derivative equal to zero:
12x^2 - 4 = 0.
Solve for x:
12x^2 = 4 → x^2 = 4/12 → x^2 = 1/3 → x = ±sqrt(1/3).
The points of inflection are x = -sqrt(1/3) and x = sqrt(1/3).
Finally, to find the absolute extrema on the interval [-2,3], we need to evaluate the function at the critical points and the endpoints of the interval.
5. Evaluate the function at the critical points and endpoints:
y(-2) = (-2)^4 - 2(-2)^2 + 1 = 16 - 8 + 1 = 9.
y(3) = (3)^4 - 2(3)^2 + 1 = 81 - 18 + 1 = 64.
y(-1) = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0.
y(1) = (1)^4 - 2(1)^2 + 1 = 1 - 2 + 1 = 0.
y(0) = (0)^4 - 2(0)^2 + 1 = 0 - 0 + 1 = 1.
The absolute extrema on the interval [-2,3] are:
Absolute maximum at (3, 64).
Absolute minimums at (-2, 9), (-1, 0), and (0, 1).
So, in summary:
Relative maximum at (0, 1).
Relative minimums at (-1, 0) and (1, 0).
Points of inflection at x = -sqrt(1/3) and x = sqrt(1/3).
Absolute maximum at (3, 64).
Absolute minimums at (-2, 9), (-1, 0), and (0, 1).