What is the value of $b$ if $-x^2+bx-5<0$ only when $x\in (-\infty, 1)\cup(5,\infty)$?
I need to find the value of $b$ that makes the quadratic inequality hold for the given intervals of $x$. I notice that the inequality is negative between the roots of the corresponding quadratic equation, and positive outside the roots. So the roots of $-x^2+bx-5=0$ must be $x=1$ and $x=5$. To find $b$, I can use the fact that the sum of the roots of a quadratic equation is equal to the opposite of the coefficient of the linear term, divided by the coefficient of the quadratic term. In this case, that means $1+5=-b/-1$, so $b=6$. I can check my answer by plugging in some values of $x$ from the intervals and seeing if the inequality holds. For example, if $x=0$, then $-x^2+bx-5=-5<0$, which is correct. If $x=3$, then $-x^2+bx-5=-9+18-5=4>0$, which is also correct. So I am confident that $b=6$ is the correct answer. 