Question

What is the domain and range of `5/(4(x+2)^2) -1`?

Answer 1

domain: `x> -2` or `x<-2` range: f(x)>-1

Explanation:

we should take a look at the denominator and notice that it resets if x=-2.

the domain: we'll compare the denominator to 0 to get `4(x+2)^2=0:` which gives us as written, x=-2. so the domain is `x> -2` or `x<-2`

for the range: we'll find the inverse function and compare it to 0. the original function: `y=5/(4(x+2)^2) -1` the inverse(by replacing x and y): `x=5/(4(y+2)^2) -1`, and we'll solve it for y, to obtain: `x \:4(y+2)^2=\frac{5}{4(y+2)^2} \:4(y+2)^2-1\cdot \:4(y+2)^2`.

now we get: `4x(y+2)^2=5-4(y+2)^2` which becomes `4(y+2)^2(x+1)=5` which becomes `(y+2)^2=\frac{5}{4(x+1)}`

we'll find the 2 roots:

for `y+2=\sqrt{\frac{5}{4(x+1)}}` we get the domain f(x)>-1
and for `y+2=-\sqrt{\frac{5}{4(x+1)}}` we get the domain f(x)>-1

so the range is f(x)>-1