A quantity z is called a functional of f(x) in the interval [a,b] if it depends on all the values of f(x) in [a,b]. What is the difference between a functional and a composite function?

Composition of functions is when you "feed" the result of one function into another function to produce yet a third function. For example, if $f(x)=x^2$ and $g(x)=e^x$ then the composition $g\circ <!--\; \circ \; -->f$ would be defined by $(g\circ <!--\; \circ \; -->f)(x)=g(f(x))=g(x^2)=e^{x^2}$. As you can see, the result is a function of $x$.
A functional, on the other hand, is when you "feed" a function -- a whole function, not just the value of the function at a specific point -- into some kind of "machine" that assigns a single numerical value to it.
For example, here are some examples of functionals:
- $F(f)={\int <!--\; \int \; -->}_0^{6}f(x)dx$. For $f(x)=x^2$, we'd have that $F(f)=72$.
- $G(f)=max\{f(x)|-<!--\; -\; -->5\le <!--\; \le \; -->x\le <!--\; \le \; -->3\}$. For $f(x)=x^2$, we'd have that $G(f)=25$.
- $H(f)=\text{the number of critical points of }f(x)\text{ on }[-<!--\; -\; -->5,3]$ . For $f(x)=x^2$, we'd have $H(f)=1$.
Notice that when you apply a functional to a function, the result is a single number. That's what is meant by the statement that the value of $F(f)$ depends, in some sense, on the "entirety" of $f(x)$ in a particular domain.
Notice also that in each of these examples the definition of the functional requires some choice of interval; different choices would lead to different results. Finally, a particular functional may only be defined for certain classes of functions; for example, neither of the examples $F$ and $G$ above are not defined for a discontinuous function with a vertical asymptote at $x=2$. So in defining a function, one usually needs to limit one's attention to some category of "nice" or "good" functions on which the functional will operate.

A functional takes a function and gives you a number.
For example the functional
$${\int <!--\; \int \; -->}_a^{b}f(x)dx$$
takes $f(x)=x^2$ and turns it into $\frac{b^3-<!--\; -\; -->a^3}{3}.$
Another functional is $z=f^{″}(0)$ which takes $f(x)=3x^2+1$ and turns it into $6$
As you see a functional is not a composite function, but it is an operator whose domain is a vector space of functions and its range is the field of that vector space.