In my book there is the following:
Although the class of primitive recursive functions contains a great many functions of practical interest, it does not include all the Turing-computable or effectively computable functions. It does not even include all the effectively computable total functions. It is therefore natural to ask how the class of primitive recursive functions can be extended so as to admit a larger number of effectively computable functions.
One approach is to remove the upper bound from the minimalization operator. The resulting composiition rule leads us to define a new class of effectively computable functions called the $\mu-$recursive functions. With the help of an important technique known as arithmetization, it is possible to show that Ackermann's functionis a $\mu-$recursive function. The same technique can also be used to show that every Turing-computable function is $\mu-$recursive and consequently that the class of $\mu-$recursive functions is identical to the class of Turing-computable functions.
Could you explain to me the part:
One approach is to remove the upper bound from the minimalization operator.
Does this mean that the primitive functions are bounded above by the minimalization operator?
If $f(x_1,x_2,\dots,x_n)$ is a primitive recursive function, then $$g(y,x_2,\dots,x_n)=\min_{k<y}f(k,x_2,\dots,x_n)=0 \;(\mbox{else }y) $$ is also a primitive function.
Hence primitive recursive functions are closed under limited minimisation. However if you remove the bound, you obtain the $\mu$ operator and get a larger class : the recursive functions.