Given f(x)=1/(`sqrtx` ), x>0, show 1/(`2(k+1)sqrt(k+1)` )<1/`sqrtk ` - 1/(`sqrt(k+1)` ) < 1/(`2ksqrtk` )?

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You need to use Lagrange's theorem such that:

`f'(c) = (f(k+1) - f(k))/(k + 1 - k) => f'(c) = (f(k+1) - f(k))`

`c in (k,k+1)`

Replacing `k+1` and `k` for `x` in equation of function yields:

`f'(c) = 1/(sqrt(k+1)) - 1/(sqrt k) => f'(c) = -(1/(sqrt k) - 1/(sqrt(k+1)))`

Replacing -`f'(c)` for `(1/(sqrt k) - 1/(sqrt(k+1)))` in inequality, yields:

`1/(2(k+1)sqrt(k+1)) < -f'(c) < 1/(2ksqrt k)`

You need to differentiate the function with respect to ` x` , such that:

`f'(x) = (1/sqrt x)' => f'(x) = -1/(2xsqrt x)`

Replacing `k` and `k+1` for `x` in equation of derivative yields:

`f'(k) = -1/(2ksqrt k)`

`f'(k+1) = -1/(2(k+1)sqrt(k+1))`

Replacing `-f'(k)` and `-f'(k+1)` for `1/(2ksqrt k)` and `1/(2(k+1)sqrt(k+1))` , yields:

`-f'(k+1) < -f'(c) < -f'(k)`

Multiplying by -1 yields:

`f'(k+1) > f'(c) > f'(k)`

Since ` k < c < k +` `` and `f'(k+1) > f'(c) > f'(k)` , you need to test if the derivative increases if `x > 0` , hence, you need to check if the second order derivative is positive, such that:

`f''(x) = (-1/2*x^(-3/2))' => f''(x) = -(1/2)*(-3/2)*x^(-5/2)`

`f''(x) = 3/(4x^2sqrtx)> 0`

Since `f''(x) > 0` . the function` f'(x)` increases over `(0,oo)` , hence, the inequality `f'(k+1) > f'(c) > f'(k)` holds.

**Hence, testing if the given inequality `1/(2(k+1)sqrt(k+1)) < 1/sqrt k - 1/(sqrt (k+1)) < 1/(2ksqrt k)` holds, using Lagrange's theorem, yields the valid statement `f'(k+1) > f'(c) > f'(k).` **

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