How do you find the domain and range of y =sqrt(x-2) / (6+x)?

1 Answer
Gió
Apr 24, 2015

You need to ensure that:
1] the denominator is different from zero, so:
6+x!=0
x!=-6
2] the argument of the square root must be bigger or equal to zero, so:
x-2>=0
x>=2
These conditions are used to avoid points or intervals where your function cannot be calculated.
Finally the domain is all real x bigger or equal to 2.

For the range I evaluated the maximum through the derivative:
y'=(10-x)/(2*(sqrt(x-2))*(x+6)^2)
The maximum is at x=10 (by setting y'=0) corresponding to y=sqrt(8)/16 giving a range for y from zero to sqrt(8)/16.

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