domain of f(x)=sqrt((1+2x)/x)
We save the definition: f(x) = 1 + 2xx
1) Let's evaluate the domain of 1 + 2xx
For the expression 1 + 2xx to exist, you must check: 1 + 2xx ≥ 0
For the expression 1 + 2xx to exist, you must check: x ≠ 0
2) Let's solve in ℝ: 1 + 2xx ≥ 0
We study the sign of the numerator
⇥The expression 1 + 2x is zero at:
⇥ x = -12
⇥Its coefficient of x, 2, being positive, it is negative before and positive after -12
We study the sign of the denominator, which must not be zero
⇥The expression x is zero at x = 0 (forbidden value), it is negative before and positive after
3) Let's draw up the sign table:
x-∞ -12 0 +∞
1 + 2x-0++
x--0+
1 + 2xx+0-❌+
Solution found: x ∈ (-∞, -12] ∪ (0, +∞)
Let's create the table of the different solutions to find the overall solution:
x-∞ -12 0 +∞
1 + 2xx ≥ 0✅✅❌❌✅
x ≠ 0✅✅✅❌✅
Domain of definition: ✅✅❌❌✅
▶Domain of definition:   x ∈ (-∞, -12] ∪ (0, +∞)