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# Difference of Squares

+1
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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.

Feb 16, 2021

#1
+1

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.

Feb 16, 2021
#2
+113739
+1

$$\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ \displaystyle\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ = \displaystyle\left(\frac{1}{b}\right)^2-\left(\frac{1}{a}\right)^2\\$$

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

Feb 16, 2021