Let , .
Then, , .
Since is continuous on and differentiable on , there exists such that
.
So, there exists such that
.
Now we prove the boundaries:
Since and the problem gives us that , the last inequality implies
.
Quite an interesting problem:
Use the inequality sqrt(3) < 1.8 to prove that 1.7 < sqrt(3) < 1.75.
The book where I found this says to use the square root function in the MVT for derivatives and use that function on the interval [3,4].
Still confused how to do this.... need help!!!
Let , .
Then, , .
Since is continuous on and differentiable on , there exists such that
.
So, there exists such that
.
Now we prove the boundaries:
Since and the problem gives us that , the last inequality implies
.
note that the information √3 < 1.8 is not needed:
from
we have:
and 12/7 is approx. 1.7142 which is certainly larger than 1.7 (or 17/10 < 12/7 since: 7*17 = 119 < 120 = 10*12).
we also have:
(as a practical matter, this means that 97/56 is a VERY good rational approximation of √3).