Question

Let f() = 3x+cosx

Evaluate f^-1(1)

Answer 1

\[
1=3x+\cos x
\]

Answer 2

How do you show it's one - to - one? Graphing ?

Answer 3

What is one to one?

Answer 4

Can't remember it entirely but it has something to with the the range I believe

Answer 5

hits the graph at most once

Answer 6

No, I know what one to one means. Are you trying to prove that \(f(x)\) is one to one?

Answer 7

ah yes lol

Answer 8

The derivative is \[
f'(x) = 3-\sin(x)
\]Since \(\min\{\sin(x)\}=-1\) then\[
\min\{f'(x)\} = 3-1=2
\]

Answer 9

oooooh

Answer 10

\[
\forall x\quad f'(x)>2\implies f(x) \text{ monotonically increasing}
\]

Answer 11

Suppose that it was not one to one. Then there must exists two distinct numbers \(a,b\) such that \(f(a)=f(b)\). But if \(a<b\) then \(f(a)<f(b)\) and if \(a>b\) then \(f(a)>f(b)\) which is a contradition.

Answer 12

So how do you actually solve this inverse? I don't think x can be solved in terms of y?

Answer 13

got ya

Answer 14

It doesn't have to necessarily be an elementary function I suppose.

Answer 15

So you're basically using inspection then lol

Answer 16

Well finding roots was never a trivial thing in mathematics.

Answer 17

You need to find the roots of \[
3x+\cos x - 1=0
\]

Answer 18

f^-1(1) = x means f(x) = 1

3x+cosx = 1

f(0)= 3(0)+cos(0) = 1

Answer 19

Yes, always try easy angles.

Answer 20

:P

Answer 21

f^-1(1) therefore = 0

Answer 22

Also take node of the amplitude of the trig function. That narrows things down a lot.

Answer 23

For example, we knew that \(x<1\) because that \(3x\) was very large.

Answer 24

mhm

Answer 25

If it wasn't 0, it was going to be a very nasty number because trig functions need transcendental numbers just to get algebraic outputs.

Answer 26

Thanks izzy :D

Answer 27

I mean wio ;)