Question

Укажіть найменьше значення виразу x²-2x+1

Answer 1

Объяснение:

$x {}^{2} - 2x + 1 = 0 \\ (x - 1) {}^{2} = 0 \\ x = 1$

получается только один корень : х=1.

Answer 2

Ответ:

Объяснение:

x²-2x+1=(х-1)²≥0 при любом х

Наименьшее значение равно 0 и достигается при х=1

Ответ:0

Answer 3

To find the minimum value of the expression x² - 2x + 1, we can use calculus. Specifically, we can take the derivative of the expression with respect to x and set it equal to zero to find the critical points. Then, we can check whether these points correspond to a minimum value.

Let's proceed with the calculation:

1. Find the derivative of x² - 2x + 1 with respect to x: d/dx(x² - 2x + 1) = 2x - 2.

2. Set the derivative equal to zero and solve for x to find the critical points: 2x - 2 = 0 2x = 2 x = 1.

The critical point is x = 1.

To determine whether it's a minimum or maximum, we can take the second derivative:

3. Find the second derivative of x² - 2x + 1 with respect to x: d²/dx²(x² - 2x + 1) = d/dx(2x - 2) = 2.

Since the second derivative is a positive constant (2), this means the original function is concave up everywhere. Therefore, the critical point at x = 1 corresponds to a minimum value.

Now, let's find the minimum value by substituting x = 1 into the original expression:

x² - 2x + 1 = 1² - 2(1) + 1 = 1 - 2 + 1 = 0.

So, the minimum value of the expression x² - 2x + 1 is 0.