Math 22 Functions

Basic Definitions

A function is a relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable.

The domain of the function is the set of all values of the independent variable for which the function is defined.

The range of the function is the set of all values taken on by the dependent variable.

Function notation: We usually denote a function f of x as $f(x)$. For example, function $y=2x^{2}+1$ can be written as $f(x)=2x^{2}+1$ in function notation.

Exercises Find the domain and range of the following functions:

1) $y={\sqrt {x+1}}$

Solution:
The domain is where the function defines (or all possible values of x). So, the radicand (everything under the square root) need to be non-negative.
So, $x+1\geq 0$
Answer: $x\geq -1$ or Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [-1,\infty )}
The range is all of possible outcomes (values of y). Notice that ${\sqrt {x+1}}$ is never negative. So $y$ is never negative.
Answer: $y\geq 0$ or $[0,\infty )$

Evaluate a Function

To evaluate a function $f(x)$ at $x=a$. We just need to plug in $x=a$ to find $f(a)$.

Example: Find the value of the function $f(x)=4x^{2}+1$ at $x=1,2,3$

Answer:

$f(1)=4(1)^{2}+1=4+1=5$

$f(2)=4(2)^{2}+1=16+1=17$

$f(3)=4(3)^{2}+1=36+1=37$

Exercises Find the value of the function at the given values:

2) $y={\sqrt {x+1}}$ at $x=3,-3$

Solution:
$f(3)={\sqrt {3+1}}={\sqrt {4}}=2$
$x=-3$ isn't in the domain of $f(x)$. So, $f(-3)=$ undefined
OR
$f(-3)={\sqrt {-3+1}}={\sqrt {-2}}=undefined$

Combinations of Functions

Two functions can be combine in varuious way. For example, let $f(x)=2x+1$ and $g(x)=x^{2}+3$. Then,

$f(x)+g(x)=(2x+1)+(x^{2}+3)=x^{2}+2x+4$

$f(x)-g(x)=(2x+1)-(x^{2}+3)=-x^{2}+2x-2$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)g(x)=(2x+1)(x^{2}+3)=2x^{3}+x^{2}+6x+3}

${\frac {f(x)}{g(x)}}={\frac {2x+1}{x^{2}+3}}$

Composite Function

Let $f$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g} be functions. The function given by $(f\circ g)(x)=f(g(x))$ is the composite function of $f$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g} .

Examples: Let $f(x)=2x+1$ and $g(x)=x^{2}+3$

So, $(f\circ g)(x)=f(g(x))=f(x^{2}+3)=2(x^{2}+3)+1=2x^{2}+7$

Exercises Given $f(x)=3x-2$ and $g(x)=2x^{2}-1$. Find each composite function below

1) $(f\circ g)(x)$

Solution:
$f(g(x))=f(2x^{2}-1)=3(2x^{2}-1)-2=6x^{2}-5$

2) $(g\circ f)(x)$

Solution:
$g(f(x))=g(3x-2)=2(3x-2)^{2}-1$

Inverse Functions

Informally, the inverse function of $f$ is another function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g} that “undoes” what $f$ has done. We usually denote Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g} as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f^{-1}}

 Formal definition of inverse function.
 Let $f$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g}
 be functions such that
 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (f\circ g)(x)=f(g(x))=x}

 and
 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (g\circ f)(x)=g(f(x))=x}

 Under these conditions, the function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g}
 is the inverse function of $f$, we denote $g=f^{-1}$

Important: The domain of $f$ must be equal to the range of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f^{-1}} , and the range of $f$ must be equal to the domain of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f^{-1}}

Exercise:

1) Show two functions Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=4x} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)={\frac {1}{4}}x} are inverses

Solution:
We want to show that these two functions satisfy $f(g(x))=x$ and $g(f(x))=x$. So
Consider $f(g(x))=f({\frac {1}{4}}x=4({\frac {1}{4}}x)=x$
and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x(x))=g(4x)={\frac {1}{4}}(4x)=x}
Hence, $f(x)=4x$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)={\frac {1}{4}}x} are inverses

2) Show two functions $f(x)={\frac {3}{2}}x+1$ and $g(x)={\frac {2}{3}}(x-1)$ are inverses

Solution:
We want to show that these two functions satisfy $f(g(x))=x$ and $g(f(x))=x$. So
Consider $f(g(x))=f({\frac {2}{3}}(x-1))={\frac {3}{2}}[{\frac {2}{3}}(x-1)]+1=(x-1)+1=x$
and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x(x))=g({\frac {3}{2}}x+1)={\frac {2}{3}}({\frac {3}{2}}x+1-1)={\frac {2}{3}}({\frac {3}{2}}x)=x}
Hence, $f(x)={\frac {3}{2}}x+1$ and $g(x)={\frac {2}{3}}(x-1)$ are inverses

Finding Inverse Function

 To find the inverse function $f^{-1}(x)$ of a given function $f(x)$. We can follow these steps:
 
 1) Replace $f(x)$ with $y$
 2) Interchange $x$ and $y$
 3) Solve for $y$
 4) Replace $y$ by $f^{-1}(x)$

Exercises Find the inverse function of

1) $f(x)=4x-1$

Solution:
Step 1: $y=4x-1$
Step 2: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=4y-1}
Step 3: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 4y=x+1}
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {x+1}{4}}}
Step 4: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f^{-1}(x)={\frac {x+1}{4}}}

2) $f(x)={\frac {3}{2}}x+1$

Solution:
Step 1: $y={\frac {3}{2}}x+1$
Step 2: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x={\frac {3}{2}}y+1}
Step 3: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {3}{2}}y=x-1}
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {3}{2}}(x-1)}
Step 4: $f^{-1}(x)={\frac {3}{2}}(x-1)$

Return to Topics Page

This page were made by Tri Phan