Prove that (68)^1/3-4

Well, this probelm requests one of Lagrange's
consequence.

We'll assign a function f(x) = cube root (x) =
x^(1/3)

The function is continuous and it could be
differentiated over the range [64 , 68].

The Lagrange's
theorem states that:

If the followings are
true

1) f(x) continuous over [64 ,
68]

2) f(x) differentiable over the range (64 ,
68)

Then there is a point c that belongs to the range [64 ,
68], such as f(68) - f(64) = f'(c)*(68 - 64).

We'll
re-write the identity f(68) - f(64) = f'(c)*(68 -
64);

(68)^(1/3) - (64)^(1/3) =
[4/3c^(2/3)]

(68)^(1/3) - 4 =
[4/3c^(2/3)]

But (68)^(1/3) - 4 < 1/12
<=> [4/3c^(2/3)] < 1/12

[4/c^(2/3)]
< 1/4 => [1/c^(2/3)] < 1/16

We know
that 64 < c < 68 => 64^(2/3) < c^(2/3) <
68^(2/3)

1/64^(2/3) > 1/c^(2/3) >
1/68^(2/3)

1/16 > 1/c^(2/3) >
1/68^(2/3)

The identity (68)^(1/3) - 4
< 1/12 is verified for af function f(x) = x^(1/3), continuous and differentiable
over the range [64 , 68].