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

1 Answer
Psykolord1989 .
Nov 24, 2017

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

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