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

1 Answer
beginning calculus
May 4, 2018

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

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