+501 If $-5\leq a \leq -1$ and $1 \leq b \leq 3$, what is the least possible value of $\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) $? Express your answer as a common fraction.
To get the minimum value, you take the smallest value for each variable. So the minimum value is (1/(-5)) + 1/1)(1/(-5) - 1/1) = 24/25.
+118690
\(\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ \displaystyle\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ = \displaystyle\left(\frac{1}{b}\right)^2-\left(\frac{1}{a}\right)^2\\\)
For the smallest value you want the absolute value of b to be as big as possible, and the absolute value of a to be as small as possible but neither can be 0
b=3, a= -1
1/9 - 1 = -8/9
