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

1 Answer
Nov 24, 2017

Answer:

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#