Find all $u$ such that neither $2u$ nor $-2u + 7$ is in the interval $(-\infty,1)$.
We use casework. Firstly assume $2u > 1$. Then $u > \frac12$. Next assume $-2u + 7 > 1$. Then $-2u > -6$ and $u < 3$. Therefore $u \in \boxed{\left(\frac 12,3\right)}$.
To find all $u$ that satisfy the given conditions, we need to determine the values of $u$ for which both $2u$ and $-2u + 7$ are not in the interval $(-\infty, 1)$.
First, let's analyze the condition $2u \notin (-\infty, 1)$. This inequality can be rewritten as follows:
$$2u \geq 1$$
By dividing both sides of the inequality by $2$, we have:
$$u \geq \frac{1}{2}$$
Next, let's analyze the condition $-2u + 7 \notin (-\infty, 1)$. This inequality can be rewritten as follows:
$$-2u + 7 \geq 1$$
By subtracting $7$ from both sides of the inequality and then dividing both sides by $-2$ (and reversing the inequality sign since we're dividing by a negative number), we obtain:
$$u \leq 3$$
Therefore, the set of all values of $u$ that satisfy neither $2u$ nor $-2u + 7$ being in the interval $(-\infty, 1)$ is given by:
$$\left[\frac{1}{2}, 3\right]$$
To find all values of $u$ that satisfy the given conditions, we need to consider each condition separately.
First, let's determine the values of $u$ for which $2u$ is in the interval $(-\infty, 1)$. We divide both sides of the inequality $2u < 1$ by $2$, which gives us $u < \frac{1}{2}$. Hence, the solution for this condition is $u < \frac{1}{2}$.
Next, let's find the values of $u$ for which $-2u + 7$ is in the interval $(-\infty, 1)$. We can rewrite the inequality $-2u + 7 < 1$ as $-2u < -6$ by subtracting $7$ from both sides. Dividing both sides by $-2$ gives us $u > 3$. So, the solution for this condition is $u > 3$.
Now, we need to find the values of $u$ that satisfy both conditions. To find the intersection of the two solution sets, we take the values of $u$ that satisfy both $u < \frac{1}{2}$ and $u > 3$. However, no value of $u$ can satisfy both conditions simultaneously because these two solution sets do not overlap. Therefore, there are no values of $u$ that satisfy neither $2u < 1$ nor $-2u + 7 < 1$.
In conclusion, there are no values of $u$ that satisfy the given conditions.