Find positive numbers M, N such that \forall x\in[-1,1]

M\leq\left|\frac{2x+7}{5-3x}\right|\leq N

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This problem requires the following fact. For a,b,c,d\in\mathbb{R}, if 0\leq a< b and 0\leq c< d then we may divide the smaller by the larger and the larger by the smaller to preserve the inequality,  i.e.

\frac{a}{d}\leq \frac{b}{c}



Now, for the upper bound, by the triangle inequality,

|2x+7|\leq |2x|+7=2|x|+7\leq 9

As the maximum of |x| as x\in[-1,1] is 1; i.e. for x\in[-1,1], |x|\leq 1. We also used |xy|=|x||y|

By the reverse triangle inequality,


For x\in[-1,1], 5-3|x| is positive [if |x|\leq 1, then 3|x|\leq 3 and so 0\leq 3-3|x|\leq 5-3|x|] so


\Rightarrow |5-3x|\geq 5-3|x|\geq 2

We have already seen 3-3|x|\geq 0; add 2 to both sides.

So we have 0\leq|2x+7|\leq9 and 0\leq 2\leq|5-3x| hence

\frac{|2x+7|}{|5-3x|}\leq \frac{9}{2}


Now, for the lower bound, by the reverse triangle inequality:

|2x+7|=|2x-(-7)|\geq \left||2x|-|-7|\right|=\left|2|x|-7\right|

For x\in[-1,1], 2|x|-7 is negative [if |x|\leq 1, then 2|x|\leq 2 and so 2|x|-7\leq -5] so


\Rightarrow |2x+7|\geq 7-2|x|\geq 2

We have already seen 2-2|x|\geq 0; add 5 to both sides.

Now by similar arguments to above:

|5-3x|\leq |5|+|-3x|=5+3|x|\leq 8

So we have 0\leq 5\leq|2x+7| and 0\leq|5-3x|\leq8 hence

\frac{5}{8}\leq \frac{|2x+7|}{|5-3x|}

Putting these together we get M=5/8 and N=9/5:

\frac{5}{8}\leq\left|\frac{2x+7}{5-3x}\right|\leq \frac{9}{5}