Functions

Functions refers to a relation between two sets that associates each element

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How to find the value and inverse of a function

For the function $\mathrm{f(x)}=\mathrm{4x\; -\; 6}$ and $\mathrm{g(x)}=\frac{\mathrm{6x\; +\; 2}}{\mathrm{2x\; +\; 3}}$, find

(a). f(4)

(b). g(2)

(c). $f^{-1}\left(x\right)$

(d). $g^{-1}\left(x\right)$

(e). $f^{-1}\left(6\right)$

(f). $g^{-1}\left(1\right)$

Solutions:

(a)

To find f(4 ) substitute 4 where there is x in f(x) = 4x - 6

$\mathrm{f(4)}=\mathrm{4(4)\; -\; 6}$

$\mathrm{f(4)}=\mathrm{16\; -\; 6}$

Answer: $\mathrm{f(4)}=\mathrm{10}$

(b)

To find g(2) we substitute 2 where there is x in $\mathrm{g(x)}=\frac{\mathrm{6x\; +\; 2}}{\mathrm{2x\; +\; 3}}$

$\mathrm{g(x)}=\frac{\mathrm{6x\; +\; 2}}{\mathrm{2x\; +\; 3}}$

$\mathrm{g(2)}=\frac{\mathrm{6(2)\; +\; 2}}{\mathrm{2(2)\; +\; 3}}$

$\mathrm{g(2)}=\frac{\mathrm{12\; +\; 2}}{\mathrm{4\; +\; 3}}$

$\mathrm{g(2)}=\frac{\mathrm{14}}{7}$

Answer: $\mathrm{g(2)}=2$

(c)

To find $f^{-1}\left(x\right)$ first equate $\mathrm{f(x)}=\mathrm{4x\; -\; 6}$ to y then make x the subject of the formula

$\mathrm{f(x)}=\mathrm{4x\; -\; 6}$

$y=\mathrm{4x\; -\; 6}$

$\mathrm{y\; +\; 6}=\mathrm{4x}$

$\frac{\mathrm{y\; +\; 6}}{4}=\frac{\mathrm{4x}}{4}$

$x=\frac{\mathrm{y\; +\; 6}}{4}$

Then where there is x write $f^{-1}\left(x\right)$ and where there is y write x

Answer: $f^{-1}\left(x\right)=\frac{\mathrm{x\; +\; 6}}{4}$

(d)

To find $g^{-1}\left(x\right)$ equate g(x) = to y and make x the subject of the formula.

$\mathrm{g(x)}=\frac{\mathrm{6x\; +\; 2}}{\mathrm{2x\; +\; 3}}$

$y=\frac{\mathrm{6x\; +\; 2}}{\mathrm{2x\; +\; 3}}$

$y\left(\mathrm{2x\; +\; 3}\right)=\mathrm{6x\; +\; 2}$

$\mathrm{6x\; +\; 2}=\mathrm{2xy\; +\; 3y}$

collect terms with x on the same side

$\mathrm{6x\; -2xy}=\mathrm{3y\; -\; 2}$

$\mathrm{x(6\; -2y)}=\mathrm{3y\; -\; 2}$

$\frac{\mathrm{x\; (6\; -2y)}}{\mathrm{(6\; -2y)}}=\frac{\mathrm{3y\; -\; 2}}{\mathrm{(6\; -2y)}}$

$x=\frac{\mathrm{3y\; -\; 2}}{\mathrm{(6\; -2y)}}$

Where there is x write $g^{-1}\left(x\right)$ where there is y write x

Answer: $g^{-1}\left(x\right)=\frac{\mathrm{3x\; -\; 2}}{\mathrm{(6\; -2x)}}$

(e)

To find $f^{-1}\left(x\right)$ substitute 6 fro x in $f^{-1}\left(x\right)=\frac{\mathrm{x\; +\; 6}}{4}$ where there is x so we have

$f^{-1}\left(x\right)=\frac{\mathrm{x\; +\; 6}}{4}$

$f^{-1}\left(6\right)=\frac{\mathrm{6\; +\; 6}}{4}$

$f^{-1}\left(6\right)=\frac{\mathrm{12}}{4}$

Answer: $f^{-1}\left(6\right)=3$

(f)

To find $g^{-1}\left(1\right)$ just substitute 1 where x is in $g^{-1}\left(x\right)=\frac{\mathrm{3x\; -\; 2}}{\mathrm{(6\; -2x)}}$

$g^{-1}\left(x\right)=\frac{\mathrm{3x\; -\; 2}}{\mathrm{(6\; -2x)}}$

$g^{-1}\left(1\right)=\frac{\mathrm{3(1)\; -\; 2}}{\mathrm{(6\; -2(1))}}$

$g^{-1}\left(1\right)=\frac{\mathrm{3\; -\; 2}}{\mathrm{(6\; -2)}}$

Answer: $g^{-1}\left(1\right)=\frac{1}{4}$

Example on how to solve a function

Solve composite functions