What is the range of the function #y=-sqrt(4x^2 + 2x^4 +5)#?

1 Answer
Jan 17, 2018

#(-infty,-sqrt(5)]#

Explanation:

So the range is all possible values of #y#, so we start to think about how low this function can go. Because the equation is negative and the argument in the square root can go to infinity, it can go forever.

The highest number will be the lowest number for the argument inside the square root. Since #x^2# and #x^4# are always non-negative, the lowest they can be is 0, so the highest value is when #x = 0#, giving the range #(-infty,-sqrt(5)]#.