Show that the linear functional $f(x) = 3\xi_1 - 5\xi_2$ is bounded on $\ell^1$ and compute its norm How do I show that $f$ is bounded and determine the norm of $f$?

Starting with,

$|f(x)| = |3\xi_1 - 5\xi_2|$ 

I am not sure what the next move is to get to the form

$|f(x)| \leq c\|x\|$ where $\|x\| = \sum_{k=1}^{\infty} |\xi_k|$
I assume $x=(\xi_i)_i$ ?

Hints: 

1) For all $a,b\in\mathbb C$ there holds $|a\pm b|\leq|a|+|b|$ and $|ab|=|a|\cdot|b|$. So, $|3\xi_1-5\xi_2|\leq \ldots$

2) $|\xi_1|\leq \|x\|$ and $|\xi_2|\leq \|x\|$
The dual space of $\ell^1$ is isometrically isomorphic to the space $\ell^\infty$. Furthermore, every bounded linear map $f : \ell^1 \to \mathbb{R}$ can be expressed as

$$f(x_n)_{n=1}^\infty = \sum_{n=1}^\infty \alpha_nx_n$$

for some sequence $(\alpha_n)_{n=1}^\infty \in \ell^\infty$, and vice-versa, every linear map of this form is in $(\ell^1)^*$. The isometrical isomorphism is given by $f \mapsto (\alpha_n)_{n=1}^\infty$, so in particular $\|f\| = \left\|(\alpha_n)_{n=1}^\infty\right\|_\infty$.

For your functional in particular we have that $f$ is represented by the vector $(3, -5, 0, 0, \ldots) \in \ell^\infty$ so we conclude that $f$ is bounded and $$\|f\| = \|(3, -5, 0, 0, \ldots)\|_\infty = \max\{|3|, |-5|\} = 5$$