Let \(p(x)=\sqrt{-x}\), and \(q(x)=8x^2+10x-3\). What is the domain of \(p(q(x))\)? Your answer will be of the form \(a\le x \le b\). Find \(b-a\).
+130085 p(x) = √ [ -x ] q(x) = 8x^2 + 10x - 3
So p(q(x) ) = √ [ - ( 8x^2 + 10x - 3 ] = √ [ 3 - 8x^2 - 10x ]
We want to find the x values that make
3 - 8x^2 -10x = 0 multiply through by -1
8x^2 + 10x - 3 = 0 factor
(4x - 1) (2x + 3) = 0
Setting both factors to 0 and solving for x we have that x = 1/4 and x = -3/2
However......the quantity under the radical, 3 - 8x^2 - 10x must be ≥ 0
So...we have three possible intervals that may solve this
(-inf, -3/2] U [ -3/2, 1/4}] U {1/4, inf )
We only need to test a point in the middle interval..if it "works," this interval will solve the equation
3 - 8x^2 - 10x ≥ 0
Testing x = 0 will make this true, so the interval that solves the equation is
-3/2 ≤ x ≤ 1/4
So b - a = 1/4 - (-3/2) = 1/4 - (-6/4) = 7 / 4
