You need to perform the first derivative test to find the maximum value of the function such that:

`f'(x) = 1 - 1/(2sqrt(7-x))`

You need to solve the equation f'(x) = 0 such that:

`1 - 1/(2sqrt(7-x)) = 0 =gt 1=1/(2sqrt(7-x)) =gt (2sqrt(7-x)) = 1`

`sqrt(7-x) = 1/2`

You need to raise to square both sides such that:

`7 -x = 1/4 =gt x = 7 - 1/4 =gt x = 27/4`

You need to use second derivative test to check if `f''(x) lt 0` such that:

`f''(x) = - 1/(2sqrt(7-x)) lt 0` , hence the critical point is a local maximum.

`f(27/4) = 27/4 + sqrt(7 - 27/4) = 27/4 + 1/2`

`f(27/4) = 29/4`

**Hence, evaluating the maximum point of the function using the first and the second derivative tests yields `(27/4;29/4).` **

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