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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 \(b-a\).

Guest Feb 10, 2018
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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

 

 

cool cool cool

CPhill  Feb 10, 2018

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