+4622 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 \(b-a\).
+9481 p( q(x) ) = √[ -(8x2 + 10x - 3) ]
To avoid the square root of a neative number, the domain must be the x values such that
-( 8x2 + 10x - 3) ≥ 0
8x2 + 10x - 3 ≤ 0
To solve this inequality, let's first find the x values that make
8x2 + 10x - 3 = 0
8x2 + 12x - 2x - 3 = 0
4x(2x + 3) - 1(2x + 3) = 0
(2x + 3)(4x - 1) = 0
2x + 3 = 0 or 4x - 1 = 0
x = -3/2 or x = 1/4
Since 8x2 + 10x - 3 is a quadratic equation, we can say that the x values that make it ≤ 0 will be either
x ≤ -3/2 and x ≥ 1/4 OR -3/2 ≤ x ≤ 1/4
If x = -2 , 8x2 + 10x - 3 = 8(-2)2 + 10(-2) - 3 = 24 - 20 - 3 = 1
So the x values in the interval x ≤ -3/2 and x ≥ 1/4 do not make 8x2 + 10x - 3 ≤ 0
If x = 0 , 8x2 + 10x - 3 = 8(0)2 + 10(0) - 3 = -3
So the x values in the interval -3/2 ≤ x ≤ 1/4 do make 8x2 + 10x - 3 ≤ 0
The domain of p( q(x) ) is -3/2 ≤ x ≤ 1/4
1/4 - -3/2 = 1/4 + 3/2 = 1/4 + 6/4 = 7/4
