Function Notation

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Question 1 of 7

1. Question
Given the function $f(x)=x+4$ , solve for:

$(i) f(1)$

$(ii) f(-3)$

$(iii) f(0)$
- $(i) f(1)=$ (5)
  
  $(ii) f(-3)=$ (1)
  
  $(iii) f(0)=$ (4)
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Function notation is a brief shortcut way to give information about a function

Simply substitute the given values as $x$

$f(1)$

$f(x)$ $=$ $x+4$

$f(1)$ $=$ $1+4$ Substitute $x=1$

$=$ $5$

$f(-3)$

$f(x)$ $=$ $x+4$

$f(-3)$ $=$ $(-3)+4$ Substitute $x=-3$

$=$ $1$

$f(0)$

$f(x)$ $=$ $x+4$

$f(0)$ $=$ $0+4$ Substitute $x=0$

$=$ $4$

$f(1)=5$

$f(-3)=1$

$f(0)=4$
Question 2 of 7

2. Question
Given the function $f(x)=2x^3-4x+3$ , solve for:

$(i) f(-3)$

$(ii) f(2)$
- $(i) f(-3)=$ (-39)
  
  $(ii) f(2)=$ (11)
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Function notation is a brief shortcut way to give information about a function

Simply substitute the given values as $x$

$f(-3)$

$f(x)$ $=$ $2x^3-4x+3$

$f(-3)$ $=$ $[2*(-3)^3]-[4*(-3)]+3$ Substitute $x=-3$

$=$ $[2*(-27)]+12+3$

$=$ $-54+15$

$=$ $-39$

$f(2)$

$f(x)$ $=$ $2x^3-4x+3$

$f(2)$ $=$ $[2*(2)^3]-[4(2)+3]$ Substitute $x=2$

$f(2)$ $=$ $(2*8)-8+3$

$f(2)$ $=$ $16-5$

$=$ $11$

$f(-3)=-39$

$f(2)=11$
Question 3 of 7

3. Question
Solve for $f(-1)-f(3)$ , given that:

$f(x)=\begin{cases}x^3-2, x≥2\\2x^2+4x-1,x<2\end{cases}$
- $f(-1)-f(3)=$ (-28)
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Function notation is a brief shortcut way to give information about a function

Piecewise Functions (aka Composite Functions) are a set of functions that share a common point

First, substitute the given values as $x$ to their respective functions

$f(x)$ $=$ $\begin{cases}x^3-2, x≥2\\2x^2+4x-1,x<2\end{cases}$

$f(-1)$

$f(x)$ $=$ $2x^2+4x-1$ Second function

$f(-1)$ $=$ $[2*(-1)^2]+[4*(-1)]-1$ Substitute $x=-1$

$=$ $2-4-1$ Evaluate

$=$ $-3$

$f(3)$

$f(x)$ $=$ $x^3-2$ First function

$f(3)$ $=$ $(3)^3-2$ Substitute $x=3$

$=$ $27-2$ Evaluate

$=$ $25$

Finally, substitute new values of the functions and solve for the final value

$\color{#00880A}{f(-1)}-\color{#9a00c7}{f(3)}$

$=$ $\color{#00880A}{-3}-\color{#9a00c7}{25}$ Substitute known values

$=$ $-28$ Evaluate

$-28$

Question 4 of 7

Solve for

$f(4)-f(0)+f(-2)$ , given that:

$f(x)=\begin{cases}3x, x≥3\\1,-1<x<3\\x^3-2,x≤-1\end{cases}$

$f(4)-f(0)+f(-2)=$ (1)

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Function notation is a brief shortcut way to give information about a function

Piecewise Functions (aka Composite Functions) are a set of functions that share a common point

First, substitute the given values as

$x$ to their respective functions

$f(x)$

$=$

$\begin{cases}3x, x≥3\\1,-1<x<3\\x^3-2,x≤-1\end{cases}$

$f(4)$

$f(x)$	$=$	$3x$	First function
$f(4)$	$=$	$3(4)$	Substitute $x=4$
	$=$	$12$	Evaluate

$f(0)$

$f(x)$	$=$	$1$	Second function
$f(0)$	$=$	$1$	Substitute $x=0$

$f(0)$

$f(x)$	$=$	$x^3-2$	Third function
$f(-2)$	$=$	$(-2)^3-2$	Substitute $x=-2$
	$=$	$-8-2$	Evaluate
	$=$	$-10$

Finally, substitute new values of the functions and solve for the final value

	$\color{#00880A}{f(4)}-\color{#9a00c7}{f(0)}+\color{#e65021}{f(-2)}$
$=$	$\color{#00880A}{12}-\color{#9a00c7}{1}+(\color{#e65021}{-10})$	Substitute known values
$=$	$1$	Evaluate

$1$

Question 5 of 7

Solve for

$f(2)-f(-2)+f(-1)$ , given that:

$f(x)=\begin{cases}2x-4, x≥1\\x+1,-1<x<1\\x^2,x≤-1\end{cases}$

$f(2)-f(-2)+f(-1)=$ (-3)

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Function notation is a brief shortcut way to give information about a function

Piecewise Functions (aka Composite Functions) are a set of functions that share a common point

First, substitute the given values as

$x$ to their respective functions

$f(x)$

$=$

$\begin{cases}2x-4, x≥1\\x+1,-1<x<1\\x^2,x≤-1\end{cases}$

$f(4)$

$f(x)$	$=$	$2x-4$	First function
$f(2)$	$=$	$2(2)-4$	Substitute $x=2$
	$=$	$4-4$	Evaluate
	$=$	$0$

$f(-2)$

$f(x)$	$=$	$x^2$	Third function
$f(-2)$	$=$	$(-2)^2$	Substitute $x=-2$
	$=$	$4$	Evaluate

$f(-1)$

$f(x)$	$=$	$x^2$	Third function
$f(-1)$	$=$	$(-1)^2$	Substitute $x=-1$
	$=$	$1$	Evaluate

Finally, substitute new values of the functions and solve for the final value

	$\color{#00880A}{f(2)}-\color{#9a00c7}{f(-2)}+\color{#e65021}{f(-1)}$
$=$	$\color{#00880A}{0}-\color{#9a00c7}{4}+\color{#e65021}{1}$	Substitute known values
$=$	$-3$	Evaluate

$-3$

Question 6 of 7

Solve for

$g(-4)+g(-2)+g(3)$ , given that:

$g(x)=\begin{cases}0, x≤1\\-5,-3<x<0\\2x,x≥0\end{cases}$

$g(-4)+g(-2)+g(3)=$ (1)

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Function notation is a brief shortcut way to give information about a function

Piecewise Functions (aka Composite Functions) are a set of functions that share a common point

First, substitute the given values as

$x$ to their respective functions

$g(x)$

$=$

$\begin{cases}0, x≤1\\-5,-3<x<0\\2x,x≥0\end{cases}$

$g(-4)$

$g(x)$	$=$	$0$	First function
$g(-4)$	$=$	$0$	Substitute $x=-4$

$g(-2)$

$g(x)$	$=$	$-5$	Second function
$g(-2)$	$=$	$-5$	Substitute $x=-2$

$g(3)$

$g(x)$	$=$	$2x$	Third function
$g(3)$	$=$	$2(3)$	Substitute $x=3$
	$=$	$6$	Evaluate

Finally, substitute new values of the functions and solve for the final value

	$\color{#00880A}{g(-4)}+\color{#9a00c7}{g(-2)}+\color{#e65021}{g(3)}$
$=$	$\color{#00880A}{0}+(\color{#9a00c7}{-5})+\color{#e65021}{6}$	Substitute known values
$=$	$1$	Evaluate

$1$

Question 7 of 7

Given the function

$f(x)=x^2+3x+5$ , solve

$\frac{f(2+h)-f(2)}{h}, h≠0$

(7+h)

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Function notation is a brief shortcut way to give information about a function

First, substitute the given values as

$x$ to the given function

$f(2+h)$

$f(x)$	$=$	$x^2+3x+5$
$f(2+h)$	$=$	$(2+h)^2+3(2+h)+5$	Substitute $x=2+h$
	$=$	$4+4h+h^2+6+3h+5$	Evaluate
	$=$	$15+7h+h^2$	Combine like terms

$f(2)$

$f(x)$	$=$	$x^2+3x+5$
$f(2+h)$	$=$	$(2)^2+3(2)+5$	Substitute $x=2$
	$=$	$4+6+5$	Evaluate
	$=$	$15$

Finally, substitute new values of the functions and solve for the final value

	$\frac{\color{#00880A}{f(2+h)}-\color{#9a00c7}{f(2)}}{h}$

$=$	$\frac{\color{#00880A}{15+7h+h^2}-\color{#9a00c7}{15}}{h}$	Substitute known values

$=$	$(7h+h^2)/h$	Evaluate

$=$	$7+h$	Simplify

$(f(2+h)-f(2))/h=7+h$

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Quiz summary

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1. Question

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2. Question

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3. Question

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4. Question

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5. Question

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6. Question

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7. Question

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Quizzes

$f(x)$	$=$	$2x^3-4x+3$
$f(-3)$	$=$	$[2(-3)^3]-[4(-3)]+3$	Substitute $x=-3$
	$=$	$[2*(-27)]+12+3$
	$=$	$-54+15$
	$=$	$-39$

$f(x)$	$=$	$2x^2+4x-1$	Second function
$f(-1)$	$=$	$[2(-1)^2]+[4(-1)]-1$	Substitute $x=-1$
	$=$	$2-4-1$	Evaluate
	$=$	$-3$

	$\color{#00880A}{f(-1)}-\color{#9a00c7}{f(3)}$
$=$	$\color{#00880A}{-3}-\color{#9a00c7}{25}$	Substitute known values
$=$	$-28$	Evaluate

$f(x)$	$=$	$x+4$
$f(1)$	$=$	$1+4$	Substitute $x=1$
	$=$	$5$