If the set $B=\{f(x) : x\in A\}$ has supremum and $C=\{k+f(x): x\in A\}$, then what is $\sup C$? 
If the set $B=\{f(x) : x\in A\}$ has supremum and $C=\{k+f(x): x\in A\}$, then what is $\sup C$?

Since $C$ is not an empty set and $f(x)\le  \sup(B) ⇒ k+f(x)\le k+\sup(B)$.
So $C$ is bounded above. Thus, $\sup(C)\le k+\sup(B)$ 
If $\sup(C)\ge k+\sup(B)$ is true, I can conclude $\sup(C)=k+\sup(B)$.
But I don't know how to show $\sup(C)\ge k+\sup(B)$.

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*$\sup(C) \geq k + \sup (B)$ is equivalent with $\sup (C) - k \geq \sup (B)$.

*Given that $\sup (C)$ is an (in fact, the smallest) upper bound on $C$, can you show that $\sup (C) - k$ is an upper bound on $B$? 

*Since $\sup(B)$ is its smallest upper bound, you'd automatically have the inequality $\sup (C) - k \geq \sup (B)$.

