Question

Найти наибольшее и наименьшее значения функции y = x + на промежутке [1, 3]

Answer 1

$y'=1-\frac{4}{x^2}\\1-\frac{4}{x^2}=0\Leftrightarrow \frac{x^2-4}{x^2}=0\Leftrightarrow \frac{(x-2)(x+2)}{x^2}=0$

Отметим знаки производной на числовой прямой:

  +           -             -            +

--------*-----------о-----------*----------->

       -2            0            2              x

Видим, что точка минимума - x = 2. Значит, на промежутке [1; 3] y(2) будет минимальным, а y(1) или y(3) - максимальным (нужно сравнить оба значения).

Наименьшее значение: 2 + 4 / 2 = 2 + 2 = 4.

y(1) = 1 + 4 = 5; y(3) = 3 + 4/3 = 4 + 1/3. y(1) > y(3).

Наибольшее значение: 5.

Ответ: y(наиб.) = 5; y(наим.) = 4

Answer 2

To find the maximum and minimum values of the function $y = x +$ on the interval $[1, 3]$, we need to evaluate the function at the endpoints of the interval and at any critical points within the interval.

Step 1: Evaluate the function at the endpoints of the interval $[1, 3]$.

• For $x = 1$, $y = 1 + 1 = 2$.
• For $x = 3$, $y = 3 + 1 = 4$.

Step 2: Find the critical points within the interval $[1, 3]$. A critical point is where the derivative of the function is equal to zero or undefined.

• The derivative of $y$ with respect to $x$ is $dy/dx = 1$.
• There are no critical points within the interval $[1, 3]$ since the derivative is a constant value (1).

Step 3: Compare the values at the endpoints and critical points (if any) to find the maximum and minimum values.

• The maximum value occurs at $x = 3$, where $y = 4$.
• The minimum value occurs at $x = 1$, where $y = 2$.

Therefore, the maximum value of the function $y = x +$ on the interval $[1, 3]$ is 4, and the minimum value is 2.