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Given y=8/(x+1) , express approx change in y in term of m when x changes from 3 to (3-m) where m is a small value.
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Simply plug 3 into y aswell as 3-m. y(3)=8/(3+1) - y(3-m)=8/((3-m)+1) 8/(3+1)-8/((3-m)+1)=?
first off \(\frac{8}{3+1}=2\)
so you are looking at \[\frac{8}{3-m+1}-2\] i.e. \[\frac{8}{4-m}-2\]
or if guess if you want "from 3 to 3 - m" you can write \[2-\frac{8}{4-m}\]and if you really want you can do the subtraction and get \[\frac{2(4-m)-8}{4-m}=\frac{8-2m-8}{4-m}=\frac{-2m}{4-m}=\frac{2m}{m-4}\]
But the answer is m/2 ??
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