What is the value of B if -x^2+bx-5<0 only when x\in (-\infty, 1)\cup(5,\infty)? I know that this is a repost but the other question did not have an answer. I have been stuck for a long time. Help Please
+128474 \( x\in (-\infty, 1)\cup(5,\infty) \)
-x^2 + bx - 5 < 0 (1) multiply though by - 1 and change the direction of the inequalitiy sign
x^2 - bx + 5 > 0 set = to 0
x^2 - bx + 5 = 0
We need to have these factors
(x - 1) ( x - 5) expanding we get that
x^2 - 6x + 5 = 0 so....this implies that we actually need
x^2 - 6x + 5 > 0 (2)
(x - 1) ( x - 5) > 0
Note that when x = ( -inf, 1) both linear factors will be negative.....so....their product will be positive
And when x = ( 5, inf) the same will occur
So multiplying (2) back through by -1 will produce (1) and we have that
-x^2 + 6x - 5 < 0 ....so......b = 6
Here's a graph to show that this function is < 0 on (-inf, 1) and (5, inf)
https://www.desmos.com/calculator/uwy0jmousq
