To solve the equation involving the Heaviside function [tex]\( H \)[/tex], we need to analyze the conditions under which the given expression holds true:
[tex]\[ H(6x + 5) - H(6x - 4) = 1 \][/tex]
The Heaviside function [tex]\( H \)[/tex] is defined such that:
- [tex]\( H(y) = 1 \)[/tex] if [tex]\( y \geq 0 \)[/tex]
- [tex]\( H(y) = 0 \)[/tex] if [tex]\( y < 0 \)[/tex]
For the difference [tex]\( H(6x + 5) - H(6x - 4) \)[/tex] to equal 1, the following conditions must be satisfied simultaneously:
1. [tex]\( H(6x + 5) = 1 \)[/tex]
2. [tex]\( H(6x - 4) = 0 \)[/tex]
Let’s analyze these conditions step by step.
Condition 1: [tex]\( H(6x + 5) = 1 \)[/tex]
For [tex]\( H(6x + 5) \)[/tex] to be 1, it must be true that:
[tex]\[ 6x + 5 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ 6x \geq -5 \][/tex]
[tex]\[ x \geq -\frac{5}{6} \][/tex]
Condition 2: [tex]\( H(6x - 4) = 0 \)[/tex]
For [tex]\( H(6x - 4) \)[/tex] to be 0, it must be true that:
[tex]\[ 6x - 4 < 0 \][/tex]
Solving this inequality:
[tex]\[ 6x < 4 \][/tex]
[tex]\[ x < \frac{2}{3} \][/tex]
Combining the Conditions
To satisfy both conditions simultaneously, [tex]\( x \)[/tex] must lie within both ranges:
[tex]\[ -\frac{5}{6} \leq x \][/tex]
and
[tex]\[ x < \frac{2}{3} \][/tex]
Combining these intervals, we get:
[tex]\[ -\frac{5}{6} \leq x < \frac{2}{3} \][/tex]
Thus, the largest possible interval in which [tex]\( x \)[/tex] could lie is:
[tex]\[ \left[ -\frac{5}{6}, \frac{2}{3} \right) \][/tex]
So, the final answer is:
[tex]\[ \boxed{\left[ -\frac{5}{6}, \frac{2}{3} \right)} \][/tex]
