How do you find the domain for #f(x) = sqrt(4 - x^5)#?

2 Answers
Alan P.
Apr 9, 2015

Assuming we are restricted to Real numbers (i.e. #f(x) epsilon RR#)
Then the domain of #f(x)=sqrt(4-x^5)#
is all values of #x# for which the argument of the square root is non-negative.
That is
#x^5<= 4#
or
#x<= root(5)(4)#
So the domain of #f(x)# is #(-oo,root(5)(4)]#

bp
Apr 9, 2015

Domain: All real numbers #<=# (4)#^1/5#

The domain of f(x) would be all real values of x for which 4-#x^5# is greater or equal to 0. The values of x which make 4-#x^5# negative cannot be accepted because in that case #sqrt(4-x^5)# would become imaginary.
Hence solving the inequality 4-#x^5# #>=# 0, it would be 4#>=# #x^5#. That is 4^#1/5# #>=# x

In the interval notation it would be ( -inf, 4^#1/5#]

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