Question

Побудуйте графік функції .y = x4 – 5x2 + 3. За графіком визначте нулі функції, точки мінімуму і максимуму, мінімальні та максимальні значення функції 2. Побудуйте графік функції y = x3 - 6x + 3.. За графіком визначте нулі функції, точки мінімуму і максимуму, мінімальні та максимальні значення функції

Answer 1

Graph of the function y = x^4 - 5x^2 + 3

To plot the graph of the function y = x^4 - 5x^2 + 3, we can start by finding the x-intercepts (zeros), minimum and maximum points, as well as the minimum and maximum values of the function.

1. Zeros of the function (x-intercepts): To find the zeros of the function, we need to solve the equation y = 0. In this case, we have the equation x^4 - 5x^2 + 3 = 0. Unfortunately, there is no simple algebraic solution to this equation. We can use numerical methods or graphing calculators to find the approximate values of the zeros.

2. Minimum and maximum points: To find the minimum and maximum points of the function, we need to find the critical points. These are the points where the derivative of the function is equal to zero or does not exist. The critical points can be found by taking the derivative of the function and setting it equal to zero.

The derivative of the function y = x^4 - 5x^2 + 3 is given by: y' = 4x^3 - 10x

Setting y' equal to zero and solving for x, we get: 4x^3 - 10x = 0 x(4x^2 - 10) = 0 x = 0 or x = ±√(10/4) = ±√(5/2)

By evaluating the second derivative at these critical points, we can determine whether they correspond to a minimum or maximum. The second derivative of the function is given by: y'' = 12x^2 - 10

Evaluating y'' at x = 0, we get: y''(0) = -10

Evaluating y'' at x = ±√(5/2), we get: y''(±√(5/2)) = 12(±√(5/2))^2 - 10 = 12(5/2) - 10 = 6

From the second derivative test, we can conclude that: - x = 0 corresponds to a relative minimum. - x = ±√(5/2) correspond to relative maximums.

3. Minimum and maximum values of the function: To find the minimum and maximum values of the function, we substitute the critical points and the endpoints of the graph into the function and compare the values.

- For the relative minimum at x = 0, the corresponding y-value is y = 3. - For the relative maximums at x = ±√(5/2), the corresponding y-values can be calculated by substituting these values into the function.

By analyzing the graph, we can determine the approximate minimum and maximum values.

Graph of the function y = x^3 - 6x + 3

To plot the graph of the function y = x^3 - 6x + 3, we can follow a similar process as before.

1. Zeros of the function (x-intercepts): To find the zeros of the function, we need to solve the equation y = 0. In this case, we have the equation x^3 - 6x + 3 = 0. Again, there is no simple algebraic solution to this equation. We can use numerical methods or graphing calculators to find the approximate values of the zeros.

2. Minimum and maximum points: To find the minimum and maximum points of the function, we need to find the critical points. These are the points where the derivative of the function is equal to zero or does not exist. The critical points can be found by taking the derivative of the function and setting it equal to zero.

The derivative of the function y = x^3 - 6x + 3 is given by: y' = 3x^2 - 6

Setting y' equal to zero and solving for x, we get: 3x^2 - 6 = 0 x^2 - 2 = 0 x = ±√2

By evaluating the second derivative at these critical points, we can determine whether they correspond to a minimum or maximum. The second derivative of the function is given by: y'' = 6x

Evaluating y'' at x = ±√2, we get: y''(±√2) = 6(±√2) = ±6√2

From the second derivative test, we can conclude that: - x = ±√2 correspond to relative minimums.

3. Minimum and maximum values of the function: To find the minimum and maximum values of the function, we substitute the critical points and the endpoints of the graph into the function and compare the values.

- For the relative minimums at x = ±√2, the corresponding y-values can be calculated by substituting these values into the function. - By analyzing the graph, we can determine the approximate minimum and maximum values.

Please note that without specific values for the x-axis range, it is difficult to provide precise values for the minimum and maximum points and the minimum and maximum values of the functions. The information provided here is based on the general behavior of the functions and the critical points obtained.