If f(x)=cosx + 3

Question

how do I find f inverse(1)?
Thanks

y = cos(x) + 3

the inverse of this is

x = cos(y) + 3

solve for y and you have your inverse

The cos function only has a range of [-1,1], so the range of f(x) is [2,4]. this means f inverse of 1 doesn't exist.

I didn't understand this.

I have to find f inverse(1) and the derivative of f inverse(1)
Answers are 0 and 1/3 respectively.

This was your question
If f(x)=cosx + 3
how do I find f inverse(1)?

Is this cos(x+3) or cos(x) + 3, there is a difference. The first one is a shift up of the cosine function. The second is a shift to the right by 3 units.
When you want to know the derivate of f inverse calculate f' , take the reciprocal and evaluate at the point.
I'm also assuming you're using radians, not degrees.
If f(x)=cos(x) + 3 then f'(x)=-sin(x)
so f'^-1(x)= -1/sin(x)

I am really really sorry.
It is f(x) = cosx + 3x

I have to find f inverse(1) and derivative of f inverse(1)

Now this is a completely different function altogether, and it's defined for all x. As x goes from (-infty,+infty) f(x) goes from (-infty,+infty).
To find f inverse 1 you want
1 = cos(x) +3x
you need some kind of root algroithm to solve this, but if you graph it you'll find x=0 then f(x)=1
f'(x)=-sin(x) + 3 so f'^-1(x) = 1/(-sin(x) + 3)
You can also see f'(0)=3 so
f'^-1(x) = 1/3

So f inverse(x) will be 1/(-sinx+3)
The derivative of f inverse(x) will be
cosx/(-sinx+3)²

That means f inverse(1) is 1/(-sin(1)+3) ? (should get 0)
And derivative of f inverse(1) is cos(1)/(-sin(1)+3)²? (ans: 1/3)

Am I doing right?

No, the derivative of the inverse function is the reciprocal of the derivative.
f inverse for this function doesn't have an elementary inverse function because of the cosine function.
When I reviewed my post I suspected there could be problems reading the text due to the font style.
Let's use an uppercase F for the function. Then we want
F'(x) and F'<su>-1(x), the derivative of the inverse function.
You want F<su>-1(1) which I said was x=0. You also wanted F'<su>-1(0), so I said to calculate F'<su>1(x) and reciprocate it.
F'<su>1(0)=3 so you should be able to see how the answer was obtained now.

I see my tags are incorrect. This is what it should be.
First, that is not how to find the inverse of a function.
No, the derivative of the inverse function is the reciprocal of the derivative.
f inverse for this function has an elementary inverse function, but it requires results you haven't had yet.
When I reviewed my post I suspected there could be problems reading the text due to the font style.
Let's use an uppercase F for the function. Then we want
F'(x) and F'^-1(x), the derivative of the inverse function.
You want F^-1(1) which I said was x=0. You also wanted F'^-1(0), so I said to calculate F'¹(x) and reciprocate it.
F'¹(0)=3 so you should be able to see how the answer was obtained now.

Once again I see an error.
Calculate F'(x) and take the reciprocal of that to find the derivative of the inverse function.