Question

$( \lbrack 0,1\rbrack \ni x: \frac{x}{1} + x = (x)f(w)$

Answer 1

$f(w) = 2$

Answer 2

The given equation is $\frac{x}{1} + x = (x)f(w)$ for $x \in [0, 1]$. The left side simplifies to $x + x = 2x$. So the equation is $2x = (x)f(w)$ for all $x \in [0, 1]$.

Let's consider the standard interpretation of juxtaposition in algebra, where $(x)f(w)$ means $x \times f(w)$. In this case, the equation is $2x = x f(w)$ for all $x \in [0, 1]$.

We can rearrange the equation as $2x - x f(w) = 0$, which is $x(2 - f(w)) = 0$. This equation must hold for all $x \in [0, 1]$.

For this product to be zero for all $x \in [0, 1]$, either $x=0$ or $2 - f(w) = 0$. If we choose any value of $x$ in the interval $(0, 1]$ (i.e., $x \neq 0$), the equation $x(2 - f(w)) = 0$ implies that $2 - f(w)$ must be zero.

$2 - f(w) = 0$

$f(w) = 2$.

This means that $f(w)$ must be equal to 2. Let's check if $f(w) = 2$ satisfies the equation for all $x \in [0, 1]$. Substitute $f(w) = 2$ into the equation $x(2 - f(w)) = 0$:

$x(2 - 2) = 0$

$x(0) = 0$

$0 = 0$.

This equation is true for all values of $x$. Since it holds for all $x \in [0, 1]$, the value $f(w) = 2$ is the solution.

Solving $2x = x f(w)$ for $f(w)$ for all $x \in [0, 1]$:

For $x \in (0, 1]$, we can divide by $x$, giving $f(w) = 2$.

For $x = 0$, the equation becomes $2(0) = 0 \cdot f(w)$, which is $0 = 0 \cdot f(w)$, or $0 = 0$. This equation is satisfied for any value of $f(w)$.

However, since the equation must hold for all $x \in [0, 1]$, the value of $f(w)$ must be the same constant that works for all $x$. The value $f(w) = 2$ works for $x \in (0, 1]$ and also works for $x=0$.

Thus, the only value of $f(w)$ that satisfies the equation for all $x \in [0, 1]$ is $f(w) = 2$.