Question

find the fixed point
f(x)=sqrt10+3x-4

Answer 1

f(x)=2+4

Answer 2

inside sqrt is just10 ?

Answer 3

no 10+3x is sqrt

Answer 4

-4 is outside

Answer 5

so than sqrt(10+3x)=4 square both sides
10+3x=16
3x=6
x=2

Answer 6

ok ?

Answer 7

let me rewrite the function

Answer 8

go

Answer 9

\[f\left( x \right)=\sqrt{10+3x}-4\]

Answer 10

does this help

Answer 11

yes than make f(x)=0

Answer 12

ok

Answer 13

and this is understandably ?

Answer 14

x=0

Answer 15

do I plug 0 in for x

Answer 16

not is zero check it what i have wrote before x=2

Answer 17

oh ok

Answer 18

now is right all ?

Answer 19

let me show you what I have

Answer 20

\[10+3x=16\]
3x=6
x=2

Answer 21

\[\sqrt{10+3(2)}-4\]\[\sqrt{10+6}-4\]\

Answer 22

\[\sqrt{16}-4\]\[\sqrt{12}\]

Answer 23

\[\approx3.46410\]

Answer 24

am I totally off the wall?

Answer 25

not is right because you have said that 4 is outside of radical

Answer 26

that is what I was thinking is what do I do with the 4

Answer 27

and sqrt16=4 because 4*4=16

Answer 28

so back where I put sqrt 16-4 iswrong

Answer 29

where is my mistake

Answer 30

yes there where you have wrote 10+3x=16 is right

Answer 31

the rest is wrong?

Answer 32

from this resulted that 3x=6 so x=2

Answer 33

at finaly

Answer 34

ok ?

Answer 35

ok

Answer 36

than good luck bye

Answer 37

bye thank you

Answer 38

yw

Answer 39

if you want to find the fixed points you need to solve \[f(x)=x\]

Answer 40

The answers are -3 and -2