Let $f(x)=1+4x-x^2, \forall x\in \mathbb{R}$
$g(x)= \max \{f(t):x \leq t \leq x+1 ; 0\leq x\leq 3\}$
I have quite basic doubt I know the meaning of $\min$ or $\max{g(x),h(x)}$ but what's the meaning of $\max{f(t)}$. It would be great if someone could explain by graph.
You are told that $g$ is a function defined on $[0,3]$.
For any $x$ in the domain of $g$ (that, is for a given $x\in[0,3]$), the value of $g(x)$ is computed by finding the largest value of $f$ on the interval $[x,x+1]$.
For example, $g(0)$ is the largest value of $1+4t-t^2$ as $t$ ranges between $0$ and $1$.
And $g(2)$ is the largest value of $1+4t-t^2$ as $t$ ranges between $2$ and $4$.
So that's why the rule for $g$ is $$g(x) = \max_{x\leq t \leq x+1} (1+4t-t^2)$$
Note: If $x\leq 1$, then $g(x)= f(x+1)$ since the max of $f$ on $[x,x+1]$ is at the right endpoint.
Similarly, if $x \geq 2$, then $g(x) = f(x)$ since the max of $f$ on $[x,x+1]$ is at the left endpoint.
And if $x\in [1,2]$, then $g(x) = f(2)$ since the max of $f$ is the global max of $f$ and this is at $2$.