Question

F(x) = √(1 - x) + √(x + 3) Найти максимум функции.

Answer 1

Ответ:

векторное решение ( без производной ) :

Пошаговое объяснение:

Answer 2

Ответ:

х = -1 - точка максимума, F(-1) = 2√2 - наибольшее значение

Пошаговое объяснение:

ОДЗ: 1 - x ≥ 0

x ≤ 1

x + 3 ≥ 0

x ≥ -3

=> -3 ≤ x ≤ 1

f'(x) = -0,5/√(1-x) + 0,5/√(x+3) = 0

√(x+3) - √(1 - x) = 0, x ≠ 1; x ≠ -3

x + 3 = 1 - x

2x = -2

x = -1 - экстремум

значения производной

___[-3]+++++++[-1]----------[1]__________

т.к. меняются с + на -, то точка максимума

F(-1) = 2√2 - наибольшее значение

F(-3) = 2

F(1) = 2

Answer 3

To find the maximum of the function F(x) = √(1 - x) + √(x + 3), we need to determine where the derivative of the function is equal to zero. Once we find the critical points, we can check which one gives the maximum value of the function.

Let's start by finding the derivative of F(x) with respect to x:

F(x) = √(1 - x) + √(x + 3)

To find the derivative, we can use the chain rule:

d/dx [√(1 - x)] = -1/2 * (1 - x)^(-1/2) * (-1) = 1 / (2 * √(1 - x))

d/dx [√(x + 3)] = 1/2 * (x + 3)^(-1/2) * 1 = 1 / (2 * √(x + 3))

Now, the derivative of F(x) is:

F'(x) = 1 / (2 * √(1 - x)) + 1 / (2 * √(x + 3))

To find the critical points, we set the derivative equal to zero and solve for x:

1 / (2 * √(1 - x)) + 1 / (2 * √(x + 3)) = 0

Next, let's manipulate the equation to get rid of the denominators:

√(x + 3) + √(1 - x) = 0

Now, move √(1 - x) to the other side:

√(x + 3) = -√(1 - x)

Now, square both sides to eliminate the square roots:

(x + 3) = (1 - x)

Now, solve for x:

2x = -2

x = -1

Now that we have the critical point x = -1, we need to verify whether this is a maximum. To do that, we can check the second derivative. If the second derivative is negative at x = -1, it is a maximum.

To find the second derivative, we differentiate the first derivative with respect to x:

d/dx [F'(x)] = d/dx [1 / (2 * √(1 - x)) + 1 / (2 * √(x + 3))]

d/dx [F'(x)] = -1 / (4 * (1 - x)^(3/2)) + 1 / (4 * (x + 3)^(3/2))

Now, substitute x = -1 into the second derivative:

d/dx [F'(x)] |x=-1 = -1 / (4 * (1 - (-1))^(3/2)) + 1 / (4 * ((-1) + 3)^(3/2)) = -1 / (4 * 2^(3/2)) + 1 / (4 * 2^(3/2)) = -1 / (4 * √2) + 1 / (4 * √2) = -1 / (4 * √2) + 1 / (4 * √2) = 0

Since the second derivative at x = -1 is zero, we cannot conclude whether it is a maximum or minimum by this method.

To finally determine whether it's a maximum, minimum, or neither, we can look at the behavior of the function around x = -1. Let's evaluate the function F(x) at points near x = -1:

F(-1.1) ≈ 1.961 F(-0.9) ≈ 2.561

Since F(-0.9) > F(-1) > F(-1.1), we can conclude that x = -1 is a local maximum. However, it is essential to note that this method doesn't guarantee it's the global maximum of the function. To find the global maximum, we would need to investigate the behavior of the function for the entire domain.