+0  
 
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1
avatar+141 

Find all real values of $p$ such that
(x+1)(x-2p)
has a minimum value of 0 over all real values of $x$.

 
 Feb 2, 2025
 #1
avatar+130313 
+1

(x + 1) (x-2p)  =

 

x^2 + (1 -2p)x -2p

 

This is a parabola that turns upward

 

The x value of the vertex =  - (1 -2p) / (2) =  -1/2 + p 

 

We want this x value to produce  0...so....

 

(x + 1) ( x -2p)  = 0

 

( -1/2 + p + 1 ) ( -1/2 + p -2p)  = 0

 

(1/2 + p) ( -1/2 -p)  = 0

 

(1/2)+ p = 0               -1/2 - p = 0

p = -1/2                      p =-1/2

 

p = -1/2

 

cool cool cool

 Feb 3, 2025

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