+0

0
57
2
+272

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
+78762
+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

CPhill  Oct 24, 2017
#2
+272
+1

Thanks so much

WhichWitchIsWhich  Oct 24, 2017

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