#f(x)=sqrt(x^4-1)+sqrt(1-x^8)# Which is set of function values? A) <0,1> B) {0} C) (0;infinite) D) (0;1>

1 Answer
George C.
Nov 10, 2017

B) #{0}#

Explanation:

Note that if #abs(x) < 1# then #x^4 < 1#. So #x^4-1 < 0# and #sqrt(x^4-1)# is undefined as a real valued function.

Also if #abs(x) > 1# then #x^8 > 1#. So #1-x^8 < 0# and #sqrt(1-x^8)# is undefined as a real valued function.

So we must have #abs(x) = 1#

In which case #x^4 = x^8 = 1# and #f(x) = 0+0 = 0#

So the domain of #f(x)# is #{1, -1}# and its range is #{0}#

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