Given y = x^4 − 4x^2 + 50. Find any extrema and any points of inflection
To find the extrema and points of inflection of the function y = x^4 − 4x^2 + 50, we need to find the first and second derivatives of the function.
First, let's find the first derivative of y with respect to x:
dy/dx = 4x^3 - 8x
To find the extrema, we can set the first derivative equal to zero and solve for x, since the critical points occur where the derivative is zero or undefined:
0 = 4x^3 - 8x
Next, let's factor out 4x:
0 = 4x(x^2 - 2)
Setting each factor equal to zero:
4x = 0 --> x = 0
(x^2 - 2) = 0
From the second equation, we find:
x^2 = 2 --> x = ± √2
So, the critical points are: x = 0, x = √2, and x = -√2.
To determine if these points are maximum or minimum points, we can use the second derivative test. We need to find the second derivative of the function, which is the derivative of the first derivative:
d²y/dx² = 12x^2 - 8
Now we can plug in the critical points into the second derivative:
At x = 0:
d²y/dx² = 12(0)^2 - 8 = -8
At x = √2:
d²y/dx² = 12(√2)^2 - 8 = 16
At x = -√2:
d²y/dx² = 12(-√2)^2 - 8 = 16
Using the second derivative test:
- If the second derivative is positive at a critical point, it is a local minimum.
- If the second derivative is negative at a critical point, it is a local maximum.
- If the second derivative is zero or undefined, the test is inconclusive.
From our calculations:
- At x = 0, d²y/dx² < 0, so this is a local maximum.
- At x = √2 and x = -√2, d²y/dx² > 0, so these are local minimums.
To find the points of inflection, we need to find the x-values where the second derivative is equal to zero or undefined:
12x^2 - 8 = 0
Simplifying the equation:
12x^2 = 8
x^2 = 8/12
x^2 = 2/3
x = ± √(2/3)
So, the points of inflection are x = √(2/3) and x = -√(2/3).
In summary:
- There is a local maximum at x = 0.
- There are two local minimums at x = √2 and x = -√2.
- There are points of inflection at x = √(2/3) and x = -√(2/3).