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 Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18 over all real values of x. (In other words, we cannot have x be nonreal.)

WhichWitchIsWhich  Oct 24, 2017
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 #1
avatar+78604 
+1

2(x+4)(x-2p) .....expanding, we have

 

2 [ x^2 + (4-2p)x -8p] 

 

2x^2 + (8-4p)x - 16p

 

And the minimum will occur where the derivative is 0

 

Take the derivative of the function and set to 0

 

4x + 8 - 4p  = 0

 

x + 2 - p  = 0

 

x = p - 2

 

So.......subbing this back into the function, we have

 

2 ( p - 2 + 4)(p - 2 - 2p)  =  -18

 

-2 ( p + 2) (p + 2)  = -18

 

(p + 2)^2  =  9     

 

p + 2  = 3              or    p + 2  = -3

 

p = 1                            p = -5

 

Here's the graph to prove this : 

 

https://www.desmos.com/calculator/6dn053wdo2

 

 

cool cool cool

CPhill  Oct 24, 2017
 #2
avatar+272 
+1

Thanks so much

WhichWitchIsWhich  Oct 24, 2017

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