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avatar+2709 

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 \(b-a\).

tertre  Feb 4, 2018
 #1
avatar+7068 
+2

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

hectictar  Feb 4, 2018
 #2
avatar+2709 
+1

Very Nice solution, hectictar! Great! smiley

tertre  Feb 5, 2018

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