Answer
(a)$$x=9$$
(b) There exists no $x$ such that $1+ \sqrt{x}=0$.
(c)$$x \ge 25$$
(d) $y=1+ \sqrt{x}$ has no maximum, and the minimum value for $y=1+ \sqrt{x}$ is $1$.
Work Step by Step
(a) Let us find the values of $x$ for which $1+ \sqrt{x}=4$; $$\Rightarrow \quad \sqrt{x}=3 \quad \Rightarrow \quad x=9.$$
(b) Let us find the values of $x$ for which $1+ \sqrt{x}=0$; $$\Rightarrow \quad \sqrt{x}=-1.$$ The above equation has no solution in the set of real numbers, so there exists no $x$ such that $1+ \sqrt{x}=0$.
(c) Let us find the values of $x$ for which $1+ \sqrt{x} \ge 6$; $$\Rightarrow \quad \sqrt{x} \ge 5 \quad \Rightarrow \quad x \ge 25.$$
(d) $y=1+ \sqrt{x}$ has no maximum since for $x_1,x_2 \ge 1$, if $x_2 > x_1$ then $y_2=1+ \sqrt{x_2} > 1+ \sqrt{x_1}=y_1$. So for $x \ge 1$, $y=1+ \sqrt{x}$ is a strictly increasing function and there is no upper bound for it.
The minimum value of $\sqrt{x}$ is $0$, so the minimum value for $y=1+ \sqrt{x}$ is $1$.
