# Why is f(n) of class O(g(n)) in this graph?

Acoording to the definition of Big-O, in this function

$$f (n) \le c g(n), \quad \text{for } n \ge n_0$$ $f (n)$ is $O(g(n))$.

But a description of a function in terms of big O notation usually only provides an upper bound on the growth rate of the function.

So for example 34 is a upper bound for the set $\{ 5, 10, 34 \}$.

So in this graph, how is $f(n)$ in $O(g(n))$ because if I get the upper bound of $g(n)$ function its value would be different than what is mentioned here for $n\ge n_0$.

I think you're confused about the notion of upper bounds for a function. We're not trying to compute an upper bound for the function $f$ in the sense of saying that, e.g., $f(n) \leq 45$ for all $n$ (or all $n$ greater than some $n_0$). Rather, we're trying to say that, for all $n>n_0$, the value (number) $c\,g(n)$ is an upper bound for the value $f(n)$. The graph indicates this informally (since it only shows what happens for some unstated but limited range of values of $n$).