If #f(x) = sqrt(x - 5)#, for #x ≥ 5#, what is the inverse of #f(2)# ?

The value at #x=2# cannot be defined for the function
#f(x)=sqrt(x-5)# because her domain is #[5,+oo)#.

It is possible to make progress if the wording of the question is interpreted in a particular way (it might be considered ambiguous in its current form).

The given function #f(x) = sqrt(x - 5)# is not one-one if the positive and negative values of the root are intended. Under those circumstances, the function would reflect in the line #y = 0#. In fact, it would be a parabola "on its side".

If it was intended by the author of the question that the function returns only the positive square root, then it will be one-one (see below). In fact, some people use the radical symbol to refer only to the positive square root (in contrast to using index notation to raise the argument to the power of one half, which would, by convention, refer to both the positive and the negative roots)

With this restriction, using the chain rule (quod vide) and, ironically but for convenience using the power notation, it might be noted

#f'(x) = 1 * 1/2 (x - 5)^(-1/2)#

returning to the radical notation with its conventional proscription for safety!

#= 1/(2 sqrt(x - 5)), x >= 5#

It might now be noted that (restricting the root to positive values)

#f'(x) >= 0# for #x >= 5#

That is, the function is a strictly rising one and therefore one-one on its specified domain.

Setting

#y = sqrt(x - 5)#

#=> y^2 = x - 5#

#=> x = y^2 + 5#

#=> f^(-1)(x) = x^2 + 5#

Next, it is possible that the phrase "the inverse of f(2)" in the question was intended to refer to "the inverse of f"(2), that is #f^(-1)(2)#, in which case,

#=> f^(-1)(2) = 2^2 + 5 = 9#