+267 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.)
+129852 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
