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

#1**+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