Functions
- Thread starter APhoenix
- Start date
A function is one-to-one if there is one range value for one domain value.
Example:
f(x)=3x+2 IS one-to-one. That is because there is only each x value results in its one y value.
\(\displaystyle f(x)=x^{2}-3\) is NOT one-to-one because x could be 1 or -1 and still result in the same value.
\(\displaystyle 1^{2}-3=-2\)
\(\displaystyle (-1)^{2}-3=-2\)
See?. Same value of -2(range) for two different inputs (domain).
If a function is one-to-one there can only be one unique value for each input.
Also, a one-to-one function has an inverse.
So, the domain of \(\displaystyle f^{-1}=\text{range of f}\)
the range of \(\displaystyle f^{-1}=\text{domain of f}\)
Example:
f(x)=3x+2 IS one-to-one. That is because there is only each x value results in its one y value.
\(\displaystyle f(x)=x^{2}-3\) is NOT one-to-one because x could be 1 or -1 and still result in the same value.
\(\displaystyle 1^{2}-3=-2\)
\(\displaystyle (-1)^{2}-3=-2\)
See?. Same value of -2(range) for two different inputs (domain).
If a function is one-to-one there can only be one unique value for each input.
Also, a one-to-one function has an inverse.
So, the domain of \(\displaystyle f^{-1}=\text{range of f}\)
the range of \(\displaystyle f^{-1}=\text{domain of f}\)
Hello, APhoenix!
\(\displaystyle f(x)\text{ is a one-to-one function if, for each value of }x,\)
. . \(\displaystyle \text{there is a }unique\text{ value of }f(x).\)
\(\displaystyle \text{Example: }\:f(x) \,=\,2x\)
\(\displaystyle \text{We see that each }x\text{ has a }unique\;f(x).\)
. . \(\displaystyle \begin{array}{cc} x & f(x) \\ \hline \vdots & \vdots \\ \text{-}2 & \text{-}4 \\ \text{-}1 & \text{-}2 \\ 0 & 0 \\ 1 & 2 \\ 2 & 4 \\ \vdots & \vdots \end{array}\)
\(\displaystyle \text{So that, if we know }f(x)\text{, then we know }x.\)
\(\displaystyle \text{If we are told that }f(x) \,=\,18,\,\text{ we know that }x = 9.\)
. . \(\displaystyle \text{That is: }\:f^{\text{-}1}(18) \,=\,9\)
\(\displaystyle \text{We are given a one-to-one function: }\,f(\text{-}3) \,=\,2.\)
. . \(\displaystyle \text{Therefore: }\:f^{\text{-}1}(2) \,=\,\text{-}3\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{Note that: }\,f(x) \,=\,x^2\,\text{ is }not\text{ one-to-one.}\)
. . \(\displaystyle \begin{array}{cc} x & f(x) \\ \hline \text{-}2 & 4 \\ \text{-}1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ \vdots & \vdots \end{array}\)
\(\displaystyle \text{Both }x = \text{-}2\text{ and }x = 2\text{ have the value }4.\)
\(\displaystyle \text{So if we're asked to find the inverse of 4, }\,f^{\text{-}1}(4)\)
. . \(\displaystyle \text{we must say: }\,f^{\text{-}1}(4)\text{ is }2\
r\:\text{-}2.\)
\(\displaystyle f(x)\text{ is a one-to-one function if, for each value of }x,\)
. . \(\displaystyle \text{there is a }unique\text{ value of }f(x).\)
\(\displaystyle \text{Example: }\:f(x) \,=\,2x\)
\(\displaystyle \text{We see that each }x\text{ has a }unique\;f(x).\)
. . \(\displaystyle \begin{array}{cc} x & f(x) \\ \hline \vdots & \vdots \\ \text{-}2 & \text{-}4 \\ \text{-}1 & \text{-}2 \\ 0 & 0 \\ 1 & 2 \\ 2 & 4 \\ \vdots & \vdots \end{array}\)
\(\displaystyle \text{So that, if we know }f(x)\text{, then we know }x.\)
\(\displaystyle \text{If we are told that }f(x) \,=\,18,\,\text{ we know that }x = 9.\)
. . \(\displaystyle \text{That is: }\:f^{\text{-}1}(18) \,=\,9\)
\(\displaystyle \text{We are given a one-to-one function: }\,f(\text{-}3) \,=\,2.\)
. . \(\displaystyle \text{Therefore: }\:f^{\text{-}1}(2) \,=\,\text{-}3\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{Note that: }\,f(x) \,=\,x^2\,\text{ is }not\text{ one-to-one.}\)
. . \(\displaystyle \begin{array}{cc} x & f(x) \\ \hline \text{-}2 & 4 \\ \text{-}1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ \vdots & \vdots \end{array}\)
\(\displaystyle \text{Both }x = \text{-}2\text{ and }x = 2\text{ have the value }4.\)
\(\displaystyle \text{So if we're asked to find the inverse of 4, }\,f^{\text{-}1}(4)\)
. . \(\displaystyle \text{we must say: }\,f^{\text{-}1}(4)\text{ is }2\
r\:\text{-}2.\)