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\(\text{Let $p(x)=\sqrt{-x}$, and $q(x)=8x^2+10x-3$. The domain of $p(q(x))$ can be written in the form $a\le x \le b$. Find $b-a$. }\)

 Jul 18, 2019
 #1
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p (q(x))  =  √ [ - ( 8x^2 + 10x - 3) ]

 

For the domain to be defined, - (8x^2 + 10x - 3 )  must be ≥ 0

 

So

 

-(8x^2 + 10x - 3)  ≥ 0

 

-8x^2 - 10x + 3 ≥ 0        

 

To solve this, set this = 0    and solve for x

 

-8x^2 - 10x + 3  = 0          multiply through by -1

 

8x^2 + 10x - 3  = 0           factor

 

(4x - 1) ( 2x + 3)   = 0

 

Set each factor to  0  and solve for x

 

4x - 1  = 0           and        2x + 3  = 0

4x  = 1                               2x   = - 3

x = 1/4                                 x  = - 3/2

 

The solution will either come from  x  = (-inf, -3/2] U [1/4, infinity)  or   x  = [-3/2, 1/4]

Testing a point in the second interval  [ I'll pick x = 0 ].....we can see that

 

√ [ - ( 8(0)^2 + 10(0) - 3) ]   =   √ [ - ( -3) ]  =  √ 3     gives us a real number

 

So....the interval that solves this is   [ -3/2, 1/4 ]

 

So  b  - a  =    1/4  - (-3/2)  =  (1/4) - (-6/4)  =  7/4

 

Here's a graph to  show that the solution interval is correct : 

https://www.desmos.com/calculator/6jhmgavfr0

 

 

cool cool cool

 Jul 18, 2019

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