How do i go about solving this? |2x−1| > √(1−x^2)

I have tried spiting to 2x−1 > √(1−x^2) and 2x−1 < -√(1−x^2) and solving for x, but the second part does not work out, or at least i cant figure it out

1 Answer
turksvids
Jan 20, 2018

#-1<=x<0# or #4/5 < x <=1#

Explanation:

#sqrt(1-x^2)# has domain #-1<=x <=1#.

#|2x-1|# can be written as a piecewise function:

#2x-1# when #x>=1/2# and #1-2x# when #x<1/2# so we're solving two equations:

#1-2x > sqrt(1-x^2)# on the interval #-1 <=x < 1/2#

and

#2x-1 >sqrt(1-x^2)# on the interval #1/2<=x<1#

For the first inequality, let's first solve the equation:

#1-2x = sqrt(1-x^2)#

Square both sides:

#1-4x+4x^2=1-x^2#

move everything to one side:

#5x^2-4x=0 rarr x(5x-4)=0 #

So #x=0# and #x=4/5# are the intersection points.

Since the inequality is only valid on #-1 <= x < 1/2#, #x=0# is the only one that matters.

So consider the intervals #-1 <=x < 0# and #0 < x < 1/2#. Testing points in those regions tells us that #1-2x > sqrt(1-x^2)# on the first interval and #1-2x < sqrt(1-x^2)# on the second interval.

So far our solution set is #-1<=x<0#.

Now let's look at the second inequality.

First we'll solve the equality for the intersection points:

#2x-1 = sqrt(1-x^2)#

square both sides:

#4x^2-4x-1=1-x^2#

moving everything to one side:

#5x^2-4x=0 rarr x(5x-4)=0#

so the intersections are #x=0# and #x=4/5#.

Since this inequality is only valid on #-1/2 <= x <=1#, we only want #x=4/5#.

So consider the intervals #1/2 <= x < 4/5# and #4/5 < x < 1#. Testing points in those regions tells us that #2x-1 < sqrt(1-x^2)# on the first interval and #2x-1 > sqrt(1-x^2)# on the second interval. Since we want the greater than part, we add #4/5 < x <=1# to our solution set.

Overall the solution set is #-1<=x<0# or #4/5 < x <=1#.

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