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 and points of inflection for the function y = x^4 - 2x^2 + 1, and the absolute extrema on the interval [-2,3], we need to find the critical points, the second derivative, and evaluate the function at these points.
1. Find the derivative of the function: Take the derivative of y = x^4 - 2x^2 + 1 with respect to x.
y' = 4x^3 - 4x
2. Set the derivative equal to zero to find the critical points:
4x^3 - 4x = 0
Factor out 4x:
4x(x^2 - 1) = 0
When 4x = 0, we get x = 0.
When x^2 - 1 = 0, we get x = -1 and x = 1.
Therefore, the critical points are x = -1, x = 0, and x = 1.
3. Find the second derivative of the function: Take the derivative of y' with respect to x.
y'' = 12x^2 - 4
4. Evaluate the second derivative at the critical points:
y''(-1) = 12(-1)^2 - 4 = 12 - 4 = 8
y''(0) = 12(0)^2 - 4 = 0 - 4 = -4
y''(1) = 12(1)^2 - 4 = 12 - 4 = 8
The second derivative is positive at x = -1 and x = 1 and negative at x = 0.
5. Determine the relative extrema and points of inflection based on the first and second derivative test:
- Relative Extrema: A relative minimum occurs when there is a change from a decreasing slope to an increasing slope. A relative maximum occurs when there is a change from an increasing slope to a decreasing slope.
- At x = -1, the second derivative y''(x) is positive, indicating a relative minimum.
- At x = 1, the second derivative y''(x) is positive, indicating a relative minimum.
Therefore, there are relative minima at (x, y) = (-1, 0) and (x, y) = (1, 0).
- Points of Inflection: A point of inflection occurs when the concavity changes (from concave up to concave down or from concave down to concave up).
- At x = 0, the second derivative y''(x) is negative, indicating a change in concavity.
Therefore, there is a point of inflection at (x, y) = (0, 1).
6. Find the absolute extrema on the interval [-2,3]:
To find the absolute extrema, we need to evaluate the function at the critical points and the endpoints of the interval.
- Evaluate the function at x = -2, x = -1, x = 0, x = 1, and x = 3:
y(-2) = (-2)^4 - 2(-2)^2 + 1 = 16 - 8 + 1 = 9
y(-1) = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0
y(0) = (0)^4 - 2(0)^2 + 1 = 0 - 0 + 1 = 1
y(1) = (1)^4 - 2(1)^2 + 1 = 1 - 2 + 1 = 0
y(3) = (3)^4 - 2(3)^2 + 1 = 81 - 18 + 1 = 64
- Compare the values to find the absolute maximum and absolute minimum:
Absolute Maximum: y(3) = 64 at x = 3
Absolute Minimum: y(-1) = 0 at x = -1
Therefore, the absolute maximum on the interval [-2,3] is 64 at x = 3, and the absolute minimum is 0 at x = -1.