SOLUTION: how do you solve this: given the quadratic function f(x)=px(x-1)-px+2, find the range of p if f(x) is always positive.

Question 451153: how do you solve this:
given the quadratic function f(x)=px(x-1)-px+2, find the range of p if f(x) is always positive.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
First off, let's put the function in standard form:
f(x) = px^2 - 2px + 2
One necessary condition for f(x)>0 for all x is that the function be concave up.
This means that the 2nd derivative must be > 0.
f'(x) = 2px - 2p
f"(x) = 2p > 0
This gives p>0.
But this is not sufficient to guarantee that f(x)>0 for all x.
The other requirement is that the function must not cross the x-axis. This is equivalent to saying there are no real zeros of the function.
The quadratic formula gives:
x = (2p +- sqrt(4p^2 - 8p))/2p
For there to be no real zeros, 4p^2 - 8p must be less than 0.
4p^2 - 8p < 0 -> 4p^2 < 8p -> p < 2.
Therefore the range of p values is:
0 < p < 2

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