Prove that (68)^1/3-4<1/12.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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].