# use model functions and show cube root (68) - 4 < 1/12

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You may use the function `f(x) = root(3) x` , and you should use Lagrange's theorem over interval [64,68] to prove the inequality such that:

`f(68) - f(64) = f'(c)(68 - 64)`

You need to evaluate f'(c), `c in (64,68), ` such that:

`f'(c) = (1/3)*c^(-2/3) =gt f'(c) = 1/(3root(3)(c^2))`

`root(3) 68 - root(3) 64 = 4/(3root(3)(c^2))`

`root(3) 68 - 4 = 4/(3root(3)(c^2))`

If `c = 64 =gt 4/(3root(3)(c^2)) = 4/(3*16) = 1/12`

`64 lt c lt 68=gt root(3) 64 lt root(3) c lt root(3) 68 =gt root(3) (64^2) lt 3root(3) (c^2) lt 3root(3) (68^2) =gt 1/(3root(3) (64^2)) gt 1/(3root(3) (c^2)gt1/( 3root(3) (68^2))`

Notice that if `c = 64 =gt 4/(3root(3) (64^2)) = 1/12`

Using Lagrange's theorem yields `4/(3root(3) (c^2)) = root(3) 68 -4` .

**Substituting `root(3) 68 - 4` for `4/(3root(3) (c^2))` and `1/12` for `4/(3root(3) (64^2))` in inequality `4/(3root(3) (64^2)) gt 4/(3root(3) (c^2)) gt4/( 3root(3) (68^2))` yields: `1/12 gt root(3) 68 - 4` .**