Question

Solve, correct to 3 d.p ------>

\[F(x)= {3 \over x}-1 ~~~~~~ x_0=1\]

Answer 1

solve
\[F(x)=0\]?

Answer 2

Yes, for \[x_0=1\]

Answer 3

why the heck don't you get x = 3 by inspection?

Answer 4

Um... Not sure if it's F(x)=0... Then 3 would be easy to find.

Answer 5

Oh, it's F^-1(x)

Answer 6

ok i would check because if you try to solve as
\[x=F(x)\] you get
\[\frac{3}{x}-1=0\]
\[\frac{3}{x}=1\]
\[3=x\] a constant funciton and so every value of x gives 3

Answer 7

\[F^{-1}(x) \]

Answer 8

You find this, and then use the x_0=1

Answer 9

then i guess we need
\[F^{-1}(x)=\frac{3}{x+1}\] first right?

Answer 10

Yes

Answer 11

then forget about it being equal to zero because the numerator is 3

Answer 12

How did you get that?

Answer 13

\[\frac{3}{x}-1=y\]
\[\frac{3}{x}=y+1\]
\[\frac{x}{3}=\frac{1}{y+1}\]
\[x=\frac{3}{y+1}\]
\[F^{-1}(x)=\frac{3}{x+1}\]

Answer 14

Ok, true.

Answer 15

i am afraid i am not being much help here because i don't really undersand the question, but there is no way that this is equal to zero

Answer 16

The answer is 1.303 if that helps in anyway?

Answer 17

You have to find the root.

Answer 18

then i really dont understand the question

Answer 19

OK. That's fine. Thanks for helping :D