Question

Y=8+2x^2-x^4найти экстремум

Answer 1

Y'=0
(8+2x^2-x^4)'=0
4x-4x^3=0
4(x-x^3)=0
x(x^2-1)=0
x=0
x=+-1

-------(-1)-------(0)------(1)----->
(-). (+). (-). (+)
xmin=-1
xmax=1

Answer 2

Finding Extremum for the Function Y=8+2x^2-x^4

To find the extremum of the function Y=8+2x^2-x^4, we can use calculus to find the critical points and determine whether they correspond to a maximum or minimum.

The first step is to find the derivative of the function Y=8+2x^2-x^4 with respect to x. The critical points occur where the derivative is equal to zero or is undefined.

The derivative of the function Y=8+2x^2-x^4 is: Y' = 0 + 4x - 4x^3

Now, let's find the critical points by setting the derivative equal to zero and solving for x: 0 + 4x - 4x^3 = 0

This equation can be factored as: 4x(1 - x^2) = 0

So, the critical points occur at x=0, x=1, and x=-1.

Next, we need to determine whether these critical points correspond to a maximum or minimum. We can use the second derivative test to determine this.

The second derivative of the function Y=8+2x^2-x^4 is: Y'' = 4 - 12x^2

Now, let's evaluate the second derivative at the critical points: - For x=0: Y''(0) = 4 > 0, so it's a local minimum. - For x=1: Y''(1) = -8 < 0, so it's a local maximum. - For x=-1: Y''(-1) = -8 < 0, so it's a local maximum.

Therefore, the function Y=8+2x^2-x^4 has local minimum at x=0 and local maximum at x=1 and x=-1.

This analysis provides the extremum points for the given function.