transformation of functions. Let $f: A \rightarrow K, f(x)=\sqrt{2 x+c}-1,$ where $A$ is the maximal domain of $f$ and $c$ is a real number.
i. For what value(s) of $c$ does $f(x) = f^{-1}(x)$ have no real roots?
ii. For what value(s) of $c$ does $f(x) = f^{-1}(x)$  have exactly one real solution?
My answers:
i)
\begin{aligned}
&\sqrt{2 x+c}-1=x\\
&2 x+c=(x+1)^{2}\\
&x^{2}+2 x+1=2 x+c\\
&\begin{array}{rl} 
& x^{2}+1-c=0 \\
a=1 & b=0 \quad c=(1-c)
\end{array}
\end{aligned}
.
.
\begin{array}{c}
D<0, b^{2}-4 a c<0 \\
0^{2}-4(1)(1-c)<0 \\
-4+4 c<0 \\
-1+c<0 \\
\therefore c<1
\end{array}
ii)
\begin{aligned}
&\begin{array}{l}
x^{2}+1-c=0 \\
b^{2}-4 a c=0 \\
0^{2}-4(1)(1-c)=0
\end{array}\\
&-4(1-c)=0\\
&\begin{array}{l}
1-c=0 \\
c=1
\end{array}
\end{aligned}
Answer is $c=1$ or $c>2$
How to do I know that $C>2$ without a calculator

You first need to find $f^{-1}(x)$, whatever the value of $c$ is. First note that the range of $f$ is $[-1,\infty)$ (since $\sqrt{2x+c}$ may attain any nonnegative value).
Here's the easy way to find $f^{-1}(x)$: from $x$ to $f(x)$ we


*

*multiply $x$ by $2$;

*add $c$

*take the square root

*subtract one


The opposite process is


*

*add one

*square

*subtract c

*divide by $2$
So $f^{-1}(x)=\frac{(x-1)^2-c}{2}$, defined for $x\geq -1$
So the equation "$f(x)=f^{-1}(x)$" actually becomes
$$\sqrt{2x+c}-1=\frac{(x-1)^2-c}{2}$$
which is what you should try to solve.
Here's an equivalent way, which might be easier: Note that "$f(x)=f^{-1}(x)$" is equivalent ot $x=f^{-1}(f^{-1}(x))$,
$$x=\frac{(\frac{(x-1)^2-c}{2}-1)^2-c}{2}$$

As a further alternative, which avoid computing $f^{-1}(x)$, you can note that "$f(x)=f^{-1}(x)$" is equivalent to $f(f(x))=x$, which becomes
$$\sqrt{2(\left(\sqrt{2x+c}-1)+c\right)}-1=x$$
