Problem 1370

1. Given the function $f(x) = ax + b$ and the condition $|f(x)| \leq 1$ for any $x \in [0,1]$, we need to find the range of values of $S = (a+1)(b+1)$.

2. Since $f(x)$ is a linear function, its maximum and minimum values on the interval $[0,1]$ will occur at the endpoints $x = 0$ and $x = 1$.

3. Evaluating $f(x)$ at the endpoints:
$$f(0) = b \quad \text{and} \quad f(1) = a + b$$
Given $|f(x)| \leq 1$, we have:
$$|f(0)| = |b| \leq 1 \quad \text{and} \quad |f(1)| = |a + b| \leq 1$$

4. These inequalities can be written as:
$$-1 \leq b \leq 1 \quad \text{and} \quad -1 \leq a + b \leq 1$$

5. To find the range of $S = (a+1)(b+1)$, we need to express $S$ in terms of $a$ and $b$ and then determine the possible values based on the constraints.

6. Expanding $S$:
$$S = (a+1)(b+1) = ab + a + b + 1$$

7. We need to find the maximum and minimum values of $S$ under the given constraints. First, consider the constraints:
$$-1 \leq b \leq 1 \quad \text{and} \quad -1 \leq a + b \leq 1$$

8. From $-1 \leq a + b \leq 1$, we can express $a$ in terms of $b$:
$$-1 - b \leq a \leq 1 - b$$

9. To find the maximum value of $S$, we need to consider the upper bounds of $a$ and $b$:
- If $b = 1$, then $-2 \leq a \leq 0$. The maximum value of $S$ occurs when $a = 0$:
$$S = (0+1)(1+1) = 1 \cdot 2 = 2$$

- If $b = -1$, then $0 \leq a \leq 2$. The maximum value of $S$ occurs when $a = 2$:
$$S = (2+1)(-1+1) = 3 \cdot 0 = 0$$

- If $b = 0$, then $-1 \leq a \leq 1$. The maximum value of $S$ occurs when $a = 1$:
$$S = (1+1)(0+1) = 2 \cdot 1 = 2$$

10. To find the minimum value of $S$, we need to consider the lower bounds of $a$ and $b$:
- If $b = 1$, then $-2 \leq a \leq 0$. The minimum value of $S$ occurs when $a = -2$:
$$S = (-2+1)(1+1) = -1 \cdot 2 = -2$$

- If $b = -1$, then $0 \leq a \leq 2$. The minimum value of $S$ occurs when $a = 0$:
$$S = (0+1)(-1+1) = 1 \cdot 0 = 0$$

- If $b = 0$, then $-1 \leq a \leq 1$. The minimum value of $S$ occurs when $a = -1$:
$$S = (-1+1)(0+1) = 0 \cdot 1 = 0$$

11. Therefore, the range of values of $S$ is from $-2$ to $2$.

The final answer is $\boxed{[-2, 2]}$.