Question

Y=5x^3-15x найти наибольшее и меньшее срочно

Answer 1

находим производную
y'=15x^2-15
y'=15(x^2-1)
приравниваем нулю производную y'=0 x^2-1=0 x1=1 x2=-1
находим y''=30x
y''(-1)=-30<0 максимум
y''(1)=30>0 имеется минимум
y(1)=5*1^3-15=-10 минимум
y(-1)=5*(-1)^3-15*(-1)=-5+15=10 максимум

Answer 2

Finding the Maximum and Minimum of Y=5x^3-15x

To find the maximum and minimum of the function Y=5x^3-15x, we can use calculus to find the critical points and determine whether they correspond to a maximum or minimum.

Step 1: Find the Derivative of Y=5x^3-15x

The derivative of Y with respect to x gives us the slope of the tangent line to the graph of Y at any point x.

The derivative of Y=5x^3-15x is given by: ``` Y' = d/dx (5x^3-15x) = 15x^2 - 15 ``` .

Step 2: Find the Critical Points

To find the critical points, we set the derivative equal to zero and solve for x: ``` 15x^2 - 15 = 0 15(x^2 - 1) = 0 x^2 - 1 = 0 (x - 1)(x + 1) = 0 ``` So, the critical points are x = 1 and x = -1.

Step 3: Determine the Nature of the Critical Points

To determine whether the critical points correspond to a maximum or minimum, we can use the second derivative test. The second derivative of Y is: ``` Y'' = d/dx (15x^2 - 15) = 30x ``` Plugging in the critical points: - When x = 1, Y''(1) = 30(1) = 30 - When x = -1, Y''(-1) = 30(-1) = -30

Since Y''(1) > 0, the critical point x = 1 corresponds to a local minimum. Since Y''(-1) < 0, the critical point x = -1 corresponds to a local maximum.

Step 4: Find the Maximum and Minimum Values of Y

To find the maximum and minimum values of Y, we substitute the critical points into the original function Y=5x^3-15x: - When x = 1, Y(1) = 5(1)^3 - 15(1) = 5 - 15 = -10 - When x = -1, Y(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10

Therefore, the minimum value of Y is -10 and the maximum value of Y is 10.

In summary, the function Y=5x^3-15x has a local minimum of -10 at x = 1 and a local maximum of 10 at x = -1.