(h/g)(-4)
\(\frac{(-4)-5}{(-4+2)(-4+3)}\)
Plug and chug
Should come out to be -4.5
\(\frac{x-5}{(x+2)(x+3)}\)
To find the values not in the domain, you must set each quantity in the denominator equal to zero and solve for x.
x+2=0 x+3=0
x≠-2 x≠-3
If x were either of these numbers, the denominator of the fraction would equal zero, which would give an undefined answer.
