Question

inverse of x^3+3x+1?

Answer 1

solve x=y^3+3y+1

Answer 2

oh also its inverse (1)...so plug in 1 im guessing?

Answer 3

f^-1(1)

Answer 4

yea but how do you solve it with two y's?

Answer 5

if it is
\[f^{-1}(1)\] you are solving
\[x^3+3x+1=1\]

Answer 6

it is very tough to solve it but if X matrix then u can solve it

Answer 7

do i do inverse though? i dont understand

Answer 8

no matrix hold on

Answer 9

if
\[f(a)=b\] then
\[f^{-1}(b)=a\]

Answer 10

so asking for
\[f^{-1}(1)\] is the same as saying "what x will give an output of 1?"

Answer 11

in other words, you know
\[f(x)=1\] you just don't know x yet

Answer 12

okay so...just tell me the first step i need to do? still a little confused

Answer 13

wouldnt x=0?

Answer 14

the first real step is to understand what
\[f^{-1}(1)\] means. it is the number you plug in to
\[f(x)\] to get 1 as an output. so your first algebra step is to write
\[f(x)=1\] i.e.
\[x^3+3x+1=1\] and solve for x

Answer 15

yes you are right

Answer 16

\[x^3+3x=0\]
\[x(x^2+3)=0\]
\[x=0\] or just do it in your head

Answer 17

ohhh okay i see it

Answer 18

i have another inverse question?

Answer 19

great. solving
\[f^{-1}(x)\] instead of
\[f^{-1}(1)\] would be more algebra

Answer 20

sure ask away

Answer 21

okay f(x)=7x/(x+1). i found the inverse to be -x/(x-7). then it says to check using function composition? how do i do (f^-1 dot f)(x) and (f dot f^-1)(x)?

Answer 22

a bunch of tedious algebra i will write the first one with explanation

Answer 23

but before i do let me check your answer because i don't want to start and end up being wrong, hold on a second

Answer 24

okay

Answer 25

yes good work! i might be easier if you wrote it as
\[f^{-1}(x)=\frac{x}{7-x}\] but maybe not. we can work with yours

Answer 26

now your job is to show that
\[(f^{-1}\circ f)(x)=x\]

Answer 27

first get rid of the circle and write what it means
\[(f^{-1}\circ f)(x)=f^{-1}(f(x))\]

Answer 28

then work from the inside out. replace the general f by the specific one you have and write
\[f^{-1}(\frac{7x}{x+1})\]

Answer 29

now compute by replacing the x in "f(x)" by 7x/(x+1)
\[f^{-1}(\frac{7x}{x+1})=\frac{-(\frac{7x}{x+1})}{\frac{7x}{x+1}-7}\]

Answer 30

yea i got this far but dont know what to do with the fractions in a fraction

Answer 31

looks nice and ugly, but since the answer will just be x you should expect a whole raft of cancelation

Answer 32

you can do it in three steps or less. first multiply numerator and denominator by x +1

Answer 33

you get
\[\frac{-7x}{7x-7(x+1)}\]

Answer 34

now you are really done. just get
\[\frac{-7x}{7x-7x-7}=\frac{-7x}{-7}=x\]

Answer 35

i couldnt cancel out 7x/x+1 before?

Answer 36

hell no!

Answer 37

like saying
\[\frac{5}{5+2}=\frac{\cancel{5}}{\cancel{5}+2}=\frac{1}{2}\]

Answer 38

haha okay. so the next equation equals x too i see. what is the range of f is my last question from this problem?

Answer 39

\[\frac{-(\frac{7x}{x+1})}{\frac{7x}{x+1}-7}\] cannot cancel anything until you write as one fraction

Answer 40

range of f is the domain of f inverse
lets me scroll up and see what f is

Answer 41

\[f(x)=\frac{7x}{x+1}\] this ratio can never be exactly 7

Answer 42

because of that 1 in the denominator

Answer 43

and you can also see that the domain of f inverse is all real numbers except 7, because you cannot divide by zero

Answer 44

ohhh okay i see! thank you i get it now

Answer 45

in other words since no input for f will give an output of 7, 7 is not in the domain of f inverse.

good luck!