Question

Знайти проміжки зростання й спадання функції та її ектремуми y=x^3-24x+2

Answer 1

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Answer 2

Finding Intervals of Growth and Decrease, and Extrema of the Function y=x^3-24x+2

To find the intervals of growth and decrease, as well as the extrema of the function y=x^3-24x+2, we can use the first derivative test. By finding the critical points and analyzing the sign of the derivative in the intervals determined by these points, we can determine where the function is increasing or decreasing and locate its extrema.

Calculating the Derivative

The first step is to calculate the derivative of the function y=x^3-24x+2. The derivative of a function gives us information about its slope and can help us identify its critical points.

The derivative of y=x^3-24x+2 is: ``` y' = 3x^2 - 24 ```

Finding Critical Points

To find the critical points, we set the derivative equal to zero and solve for x: ``` 3x^2 - 24 = 0 3x^2 = 24 x^2 = 8 x = ±√8 ```

So, the critical points are x = √8 and x = -√8.

Analyzing the Sign of the Derivative

Now, we can analyze the sign of the derivative in the intervals determined by the critical points.

1. Interval (-∞, -√8): - Pick a test point, e.g., x = -1. - Substituting x = -1 into the derivative, we get: ``` 3(-1)^2 - 24 = -21 ``` - The derivative is negative in this interval, so the function is decreasing.

2. Interval (-√8, √8): - Pick a test point, e.g., x = 0. - Substituting x = 0 into the derivative, we get: ``` 3(0)^2 - 24 = -24 ``` - The derivative is negative in this interval, so the function is decreasing.

3. Interval (√8, +∞): - Pick a test point, e.g., x = 1. - Substituting x = 1 into the derivative, we get: ``` 3(1)^2 - 24 = -21 ``` - The derivative is negative in this interval, so the function is decreasing.

Identifying Extrema

Since the function is decreasing in all intervals, it has no local maxima or minima. However, it does have an absolute minimum at the endpoint of the interval. To find the absolute minimum, we can evaluate the function at the endpoints of the intervals.

- At x = -√8, y = (-√8)^3 - 24(-√8) + 2 - At x = √8, y = (√8)^3 - 24(√8) + 2

By comparing the values of y at these points, we can determine the absolute minimum of the function.

Conclusion

The function y=x^3-24x+2 has intervals of decrease in the entire domain and no local extrema. It has an absolute minimum at one of the endpoints of the domain.