[tex]f(x) = \frac{10x + 1}{8x + 3}[/tex]
8x + 3 ≠ 0
8x ≠ -3
[tex]\frac{8x}{8} \neq \frac{-3}{8}[/tex]
x ≠ [tex]\frac{-3}{8}[/tex]
Interval Notation: (∞, [tex]\frac{-3}{8}[/tex]) U ( [tex]\frac{-3}{8}[/tex], ∞)
(- ∞, -[tex]\frac{3}{8}[/tex]) ∪ (- [tex]\frac{3}{8}[/tex], ∞ )
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
solve : 8x + 3 = 0 ⇒ x = - [tex]\frac{3}{8}[/tex]
