Find the largest δ such that if 0 < |x – 1| < δ then |f(x) – 1| < 0.1?
\[f(x) = 2- \frac{ 1 }{ x }\]
what have you worked out so far?
@amistre64 I did f(x)=2-1/x absolute value of 2-1/x < 0.1 0< absolute value of x-1< delta 1- 1/x < 0.1 Here's where I got stuck. Di I do it wrong?
we have to solve these two inequalities: \[\Large \left| {1 - \frac{1}{x}} \right| < \frac{1}{{10}}\]
your starting fine, the idea is to convert the range into the domain, which is just algebra worked over ... \[|f(x)-1|<.1\] \[-.1<f(x)-1<.1\] \[.9<f(x)<1.1\] \[.9<2 - \frac 1x<1.1\] \[.9-2<- \frac 1x<1.1-2\] \[2-.9> \frac 1x>2-1.1\] \[\frac1{2-.9}< x<\frac1{2-1.1}\] \[\frac1{2-.9}-1< x-1<\frac1{2-1.1}-1\]
we have 2 deltas to try .... test them out to see what fits.
|dw:1431802223438:dw|
the smaller value is the safest bet to me ...
I've gotten -0.09 <x-1<0.11
-1/11 < x < 1/9
o sorry
\[2-\frac1{1-\frac1{11}}=1-.1\] \[2-\frac1{1+\frac1{9}}=1+.1\]
but since 1/9 is greater than 1/11 we cant use x-1/9 and still be in our range of .9 to 1.1
\[2-\frac1{1-\frac1{9}}=1-.125\] we are out of the bounds of our limitation
That'd give us 0.875 or 7/8
correct, which is smaller than .9 we need to remain within (.9, 1.1) .875 < (.9, 1.1)
right
if we use the delta = 1/11 then the function fits in both cases. so its the largest value we can use.
which gives us 0.0909....
yes if they want a decimal representation of it
I checked it in my assignment online for that question, and 1/11 is the correct answer.
They take integers, fractions or decimals
well of course it is .. :)
@amistre64 thank you for the walk through. I just got stuck whiles working it out :)
youre welcome, ita just an exercise in algebra, converting range to the domain and picking the smaller result
u sir just got yourself a medal ;)
good luck :)
thank u
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