Answer:
Step-by-step explanation:
The domain restrictions of this expression stem from the denominator, g^2−2g−24. Remembering that we can NOT divide by zero, we purposely set g^2−2g−24 = 0 and solve for g; the resulting g values are the ones g cannot have here.
g^2−2g−24 = (g-6)(g+4) = 0. Then the forbidden g values are g = 6 and g = -4.
We can write the domain in set notation as: (-∞,-4) ∪ (-4,6) ∪ (6, infinity). Or
we can just type out the domain restrictions: g≠6 and g≠−4.
(-infinity, -4)∪
Answer:
g ≠ -4 and g ≠ 6
Step-by-step explanation:
Let,
[tex]h(g)=\frac{g^2+7g+12}{g^2-2g-24}[/tex]
Which is a rational function,
Since, a rational function is defined for all real numbers except those for which denominator = 0,
If [tex]g^2 - 2g - 24=0[/tex]
By the middle term splitting,
[tex]g^2 - (6g - 4g) - 24=0[/tex]
[tex]g^2-6g+4g - 24=0[/tex]
[tex]g(g-6)+4(g-6)=0[/tex]
[tex](g+4)(g-6)=0[/tex]
By the zero product property,
g + 4 = 0 or g - 6 = 0
⇒ g = -4 or g = 6
⇒ h(g) is defined for all real numbers except g = -4 and g = 6,
i.e. domain restriction is,
g ≠ -4 and g ≠ 6
