Learn how to do composite functions & fog x & fx functions, Practice to finding the composition of functions when two functions are given in form of tables.

To find \(fog(x)\) at a particular value of \(x\), say \(x=a\), put \(x=a\), evaluate \(g(a)\) and then put this value in the expression of \(f\).

A \(5\)

B \(\sqrt 7\)

C \(-1\)

D \(\sqrt 2\)

- If values of two function \(f\) and \(g\) are given in tabular form for some values of \(x\), we can find the value of composite of these two function at a particular value of \(x\).

e.g.

Consider the value of \(fog(8)=f(g(8))\)

\(\Rightarrow f(7)=6\)

- Sometimes it is very useful if we know how to find the component function given a complicated composite function. In other words we should be able to find \(f\) and \(g\) by looking at \(fog \).

A \(f(x)=\dfrac {x}{1+x}\) and \(g(x)=sin\,x\)

B \(f(x)=\dfrac {1+x^2}{x}\) and \(g(x)=cos\,x\)

C \(f(x)=cos\,x\) and \(g(x)=x^3+3\)

D \(f(x)=\dfrac {1}{x}\) and \(g(x)=cos\,x\)

- Consider two function \(f(x)\) and \(g(x)\), we define composite function of \('f'\) and \('g'\) as \(fog(x)=f(g(x))\) (This is different from \((fg)(x)\)).
- First \('g'\) is applied to \(x\) and then \('f'\) rule is applied to \(g(x)\) .

- The output of \('g'\) machine acts as an input for \('f'\) machine and final output is \(f(g(x))\).
- Similarly we can define \(g(f(x))=gof(x)\) or \(fof(x)=f(f(x))\)

- Similarly the composite of three function \(f, \,g, \,h\) is defined as \(fogoh(x)=f(g(h(x))\)

- Given the graph of two functions \(f\) and \(g\) if we need to find out the value of composite function of \(f\) and \(g\) at some value of \(x\) say \(f(g(\alpha))\), we find \(g(\alpha)\) first by noting the height of graph of \(g\) above the point \((\alpha, 0)\) and then the height of graph of \(f\) above the point \((g(\alpha), 0)\).
- e.g.

- The symbol \(f\,o\,g\,\,o\,h=f(g(h(x)))\).
- We evaluate \(h(x)\) first, then put the value in the rule of \(g\) to get \(g(h(x))\) and then put this value in the rule of \(f\) to get \(f[g(h(x))]\)

A \(\dfrac {2x}{x^4+1}\)

B \(\dfrac {x^4}{2x+1}\)

C \(\dfrac {2(x+2)^4}{(x+2)^4+1}\)

D \(x^5\)

- If a composite function expression of more than two functions is given then the component functions can be found by inspection.

e.g.

Let \(f(g(h(x))=sin(cos\,x^2)\), then we observe that the rightmost expression is \(x^2\) so \(h(x)=x^2\), to the left of it is \(cos\) so \(g(x)=cos\,x\) and leftmost function is \(sin\). So \(f(x)=sin\,x\)

A \(f(x)=x^4,\;g(x)=sin\,x, \;h(x)=\sqrt x\)

B \(f(x)=cos\,x,\;g(x)=x^2, \;h(x)=\dfrac {1}{x}\)

C \(f(x)=x+1,\;g(x)=tan\,x,\; h(x)=x^3\)

D \(f(x)=\sqrt [3] x,\;g(x)=cos\,x, \;h(x)=\dfrac {1}{x}\)

- Sometimes we are given the composite function expression of two functions \(f\) and \(g\) i.e. \(gof(x)\) and also the expression for \(f(x)\).To find \(g(x)\) in such case,put \(f(x)=t\) and express the \(gof(x)\) expression in terms of \(t\).

e.g.

\(f(g(x))=5x+3\) and \(g(x)=2x+1\), then \(f(2x+1)=5x+3\)

Put \(2x+1=t\Rightarrow x=\dfrac {t-1}{2}\)

\(\therefore f(t)=5\dfrac {(t-1)}{2}+3=\dfrac {5t-5+6}{2}\)

\(\Rightarrow f(t)=\dfrac {5t+1}{2}\Rightarrow f(x)=\dfrac {5x+1}{2}\)

A \(g(x)=3x+1\)

B \(g(x)=4x^2+7\)

C \(g(x)=sin\,x+2\)

D \(g(x)=\dfrac {1}{x}\)