Question

The domain for f(x)=x^3-x is all real numbers. But why? Can someone explain to me why it is all real numbers?

Answer 1

Polynomials have domain all real numbers because you can plug in any real number in the function value will exist.

Answer 2

You can also look at a doodle of the graph an observe the domain to be all real numbers.

Answer 3

Oh okay I get it now. Thank you so much!

Answer 4

np

if you had
\[f(x)=\sqrt{x^3-x}\]
you would have some restrictions on your domain

Answer 5

because sqrt( ) isn't going to have a function value everywhere per plug in

Answer 6

Oh okay so the restrictions for that problem would be x cannot equal 1,0,-1. Right?

Answer 7

\[f(x)=\sqrt{x^3-x}=\sqrt{x(x^2-1)}\]


You want the inside to be positive or zero. x(x^2-1) cannot be negative

so let's look at this...
x(x^2-1)=0 when x=0 or x=-1 or x=1
these numbers are included because the inside of the square root is 0 when x equals any of those values and sqrt(0)=0

but what about all the other numbers in the real number universe

I like to draw a number line
plot the numbers where my function is 0
then test the intervals by choosing a number representative per interval


So I need to choose a number before -1. Example x=-2 .
I need to choose a number between -1 and 0. Example x=-1/2 .
I need to choose a number between 0 and 1. Example x=1/2.
I need to choose a number after 1. Example x=2.

I will plug these into x^3-x to see if x^3-x is positive or negative.
I will throw the interval out if it's number representative me a negative result.

Answer 8

\[x^3-x \\ x=-2 \text{ gives } (-2)^3-(-2)^2=-8+2=-6 \\ \text{ this is no good because we have } \sqrt{-6} \\ \text{ so } (-\infty,-1) \text{ is not part of the domain }\]

\[x^3-x \\ x=-\frac{1}{2} \text{ gives } (\frac{-1}{2})^3-(\frac{-1}{2})= \frac{-1}{8}+\frac{1}{2}=\frac{-1}{8}+\frac{4}{8}=\frac{3}{8} \\ \text{ we do have that } \sqrt{\frac{3}{8} } \text{ exists so this interval } [-1,0] \text{ is } \\ \text {definitely part of our domain }\]


\[x^3-x \\ x=\frac{1}{2} \text{ gives } (\frac{1}{2})^3-\frac{1}{2}=\frac{1}{8}-\frac{1}{2}=\frac{1}{8}-\frac{4}{8}=\frac{-3}{8} \\ \text{ again this is no good because we have } \sqrt{\text{negative number }} \\ \text{ so we will not include } (0,1) \text{ in our domain } \]

\[x^3-x \\ x=2 \text{ gives } 2^3-2=8-2=6 \\ \text{ since } \sqrt{6} \text{ does exist } \\ \text{ then we have } [1,\infty) \text{ is part of our domain }\]

Answer 9

\[\text{ so if our function was } f(x)=\sqrt{x^3-x }\\ \text{ then our domain would be } [-1,0] \cup [1,\infty)\]

Answer 10

Ohh okay so x would be x is greater or equal to -1. Right? How would you write out the domain with greater than or less than signs?

Answer 11

so that first interval I wrote says x is between -1 and 0 inclusive
so the first inequality would look like \[-1 \le x \le 0\]
the last interval says x is greater than or equal to 1 (since it goes to infinity we don't have to say between anything)
so the last interval would look like \[x \ge 1\]

Answer 12

so if you wanted the answer to
find the domain of f(x)=sqrt(x^3-x) in the real number universe
then the answer would be \[-1 \le x \le 0 \cup x \ge 1 \]

Answer 13

you could replace that one symbol with "or" (talking about that funny U looking thing)

Answer 14

Oh okay. I get it now. Thank you so much!

Answer 15

Np
I didn't know I chose a complicated example or not. :p

Answer 16

It all good. It really helped.

Answer 17

cool stuff

Answer 18

I have another question. What would the reciprocal of 1+i be?

Answer 19

\[\text{ the reciprocal of } a \text{ is } \frac{1}{a}\]
But they are probably going to want you to write your reciprocal in standard form of a complex number. That is c+di form.

Answer 20

\[\text{ the reciprocal of } a+bi \text{ is } \frac{1}{a+bi}\]
you would need to multiply bottom and top by bottom's conjugate

Answer 21

\[\frac{1}{a+bi} \cdot \frac{a-bi}{a-bi}=\frac{a-bi}{a^2-b^2i^2}=\frac{a-bi}{a^2-b^2(-1)}=\frac{a-bi}{a^2+b^2}\]

\[\text{ the reciprocal of } a+bi \text{ is } \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i\]

Answer 22

so let's look at your problem

Answer 23

The question is write this in standard form:
\[\frac{1}{1+i}\]
What do you think you need to multiply top and bottom by?

Answer 24

By 1-i

Answer 25

correct

Answer 26

\[\frac{1}{1+i} \cdot \frac{1-i}{1-i}=\frac{1-i}{(1+i)(1-i)}\]

Answer 27

you could use "foil" on bottom but you can short cut this since you are multiplying conjugates

Answer 28

that is you just need to multiply first and last

Answer 29

"foil" method minus the "oi"

Answer 30

but you can foil it if you want

Answer 31

So the answer would be \[\frac{ 1-i }{ 2 }\]

Answer 32

right
that would be an okay answer to me
but you can also write as two separate fractions

\[\frac{1}{2}-\frac{i}{2} \text{ or } \frac{1}{2}-\frac{1}{2}i \text{ or } \frac{1}{2}+\frac{-1}{2}i\]

Answer 33

in case you have a multiple choice test you need to be able to see your answer as their answer
because sometimes our answers don't look the same but they may be the same

Answer 34

okay, thank you!

Answer 35

np