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

1 Answer

Answer:

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