I am currently doing a math problem and have come across an unfamiliar notation. A mini circle between $f$ and $h(x)$

The question ask me to find for 'the functions $f(x)=2x-1$ and $h(x)=3x+2$'

$$f \circ h(x)$$

However, I can't do this as I do not know what the circle notation denotes to. Does it mean to multiply?

  • 6
    $\begingroup$ It is called composition and is defined by $(f \circ h)(x) = f(h(x))$. $\endgroup$
    – copper.hat
    Dec 8, 2014 at 7:35

2 Answers 2


This notation means that you take the output of $h$ and use it as the input of $f$. When we are working with a specific $x$ value, we can suggestively write $f(h(x))$ instead.

For instance if $f(z)=1/z$ and $h(x)=2+3x$ then $$(f\circ h)(x) = f\big(h(x)\big) = f(2+3x) = \frac{1}{2+3x}.$$

(Note: I only used $z$ as the variable for $f$ to avoid confusion; in practice the function does not care what its input variable is named.)


The circle $\circ$ is the symbol for composition of functions. In General, if you have two functions $g\colon X\rightarrow Y$ and $f\colon Y\rightarrow Z$, then $f\circ g$ is a function from $X$ to $Z$. For $x\in X$ one has $(f\circ g)(x) = f(g(x))$.

In your case one has: $f(x) = 2x-1$, $g(x) = 3x+2$ and $$ (f\circ g)(x) = f(g(x)) = 2(g(x))-1 = 2(3x+2) -1 = 6x+3. $$ You take the function $g(x)$ and put it in place of the $x$ in the function $f$.

This is obviously different from $f(x)\cdot g(x) = (2x-1)\cdot (3x+2) = 6x^2+x-2$.


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