elementary calculus please...?

Question

Prove that (1/9) is less than or equal to SQRT(66) -8 which is less than or equal to (1/8)

(without computing SQRT66 to 2 decimal places!)

Hint: Recall the Mean Value Theorem.

Can someone please show me how to do this problem without using "integral" or integration... just using elementary calculus i guess....


thank you!

Answer 1

It is obvious that
66 - 17/81 < 66 < 66 + 1/64
is true.
It follows that
(66 - 17/81)^0.5 < 66^0.5 < (66 + 1/64)^0,5
(66 - 17/81)^0.5 - 8 < 66^0.5 – 8 < (66 + 1/64)^0,5 – 8
(5329/81)^0.5 - 8 < 66^0.5 – 8 < (4225/64)^0,5 – 8
73/9 - 8 < 66^0.5 – 8 < 65/8 – 8
1/9 + 8 - 8 < 66^0.5 – 8 < 1/8 + 8 – 8
1/9 < 66^0.5 – 8 < 1/8
Q.E.D.
-

Answer 2

√(5329/81) = √(65 64/81) < √66 < √(66 1/64) = √(4225/64)
8 1/9 < √66 < 8 1/8
1/9 < √66 - 8 < 1/8

It's not what you wanted, though.