Question

How do you find the domain and range of `f(x)= -2 + sqrt(16 - x^2)`?

Answer 1

Find the highest and lowest x-values for which the function is defined, then find the highest y-value (which will occur at `x=0` because that's where `16-x^2` is at its largest), and the lowest y-value (occurs where `16-x^2=0`.

Domain: `-4<=x<=4`
Range: `-2<=f(x)<=2`

Explanation:

Look for where the function is defined. Recall that we cannot take the square root of a negative number; thus, for all x, `16-x^2` must be positive. This will clearly only occur for `x^2<16`, aka `absx<4`

Thus, this function is only defined for `-4<=x<=4`. That is our domain.

The highest `f(x)` value will occur when `sqrt(16-x^2)` is at its highest, which will occur at `x=0`. At that point, `f(0) = -2+sqrt16 = -2+4 = 2`

The lowest `f(x)` value occurs when `sqrt(16-x^2) = 0`. At this point we have `f(4) = -2+sqrt(16-16) = -2`

Thus, our range will be `-2<=f(x)<=2`