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
The opposite process is
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 $$\sqrt{2(\left(\sqrt{2x+c}-1)+c\right)}-1=x$$