1. g/f(x)...

--->Question:
f(x) = sqrt(4*x+5)
g(x) = (x-2)/x

find g/f(x) and it's domain

-find domain for both f(x) and g(x)
f(x)-- sqrt(4*x+5) ---- under square root has to be greater than or = to zero
4x+5 >= 0
4x >= -5
x >= -5/4

g(x) -- the denominator cannot be zero
x =! 0

so for (g/f)(x), g has to be defined, f has to be defined and f can't be zero

((x-2)/x) * (1/(sqrt(4x+5)))
x =! 0 & x > -5/4 [[[[no longer greater than or equal to 0]]]]]

so domain in interval notation is
(-5/4,0)U(0,infinity)

DID I DO THIS RIGHT!?!?

2. Originally Posted by williamb
--->Question:
f(x) = sqrt(4*x+5)
g(x) = (x-2)/x

find g/f(x) and it's domain

-find domain for both f(x) and g(x)
f(x)-- sqrt(4*x+5) ---- under square root has to be greater than or = to zero
4x+5 >= 0
4x >= -5
x >= -5/4

g(x) -- the denominator cannot be zero
x =! 0

so for (g/f)(x), g has to be defined, f has to be defined and f can't be zero

((x-2)/x) * (1/(sqrt(4x+5)))
x =! 0 & x > -5/4 [[[[no longer greater than or equal to 0]]]]]

so domain in interval notation is
(-5/4,0)U(0,infinity)

DID I DO THIS RIGHT!?!?
The function $\frac{g}{f}(x) = \frac{\frac{x-2}{x}}{\sqrt{4x+5}}$

Let's re-write that:

$\frac{g}{f}(x) = \frac{x-2}{x} \times \frac{1}{\sqrt{4x+5}}$

$= \frac{x-2}{x\sqrt{4x+5}}$

Now, we need to consider which values this is defined for!

The square root on the bottom must NOT be negative OR zero. In other words, it's positive.

$4x+5 > 0$

$4x > -5$

$x > \frac{-5}{4}$

The other x on the bottom cannot be zero, so $x \neq 0$

Hence the domain is $\bigg(\frac{-5}{4}, 0\bigg) \cup \bigg(0, \infty\bigg)$