inverse

If f(x)=cosx + 3
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.

asked by Jen

Respond to this Question

First Name

Your Answer

Similar Questions

  1. Math

    y=arccos(sin(x)), find dy/dx and sketch it's graph(I guess I can do this on wolframalpha after I'm done solving the question). And by arcsin I mean inverse of the expression(written like cos^-1(sin(x)), but is not 1/cos(sinx) but
  2. Precalculus

    Find the exact value of: Cos(inverse tan 4/3 + inverse cos 5/13)
  3. Math

    Determine whether each equation is true for all x for which both sides of the equation are defined. If it is true, support your conclusion with a sketch using the unit circle. If it is false, give a counterexample. inverse
  4. Trig

    Solve for t algebraically: inverse cos(t) = inverse sin(t). Where do I start?
  5. Trigonometry

    Write equivalent equations in the form of inverse functions for a.)x=y+cos è b.)cosy=x^2 (can you show how you would solve) a.) x= y+ cos è cos è = x-y theta = cos^-1(x-y) b.) cosy=x^2 cos(y) = x^2 y = Cos^-1(x^2)
  6. inverse functions

    1-3cos=sin squared You need to define the variables, such as x, y etc. If you are trying to solve 1 - 3 cos x = sin^2 x for whatever x is, then you are not talking about an inverse function.
  7. Trig

    Compute inverse functions to four significant digits. cos^2x=3-5cosx rewrite it as.. cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos x using the equation cosx=(-5 -+ sqrt (25 +12)/2
  8. find dy/ds

    y = s*square root of(1-s^2) + cos inverse(s) Just give me some hints and I will do it. Thanks. You have y = s*square root of(1-s^2) + cos inverse(s) which I would write as y = s*sqrt(1-s2) + cos-1(s) and you want dy/ds For the
  9. Calculus (Derivatives of Inverse Functions)

    Suppose f(x) = sin(pi cos(x)). On any interval where the inverse function y = f^-1(x) exists, the derivative of f^-1(x) with respect to x is: I've come as far as y = arccos ((arcsin(x))/pi), but I am not certain this is right.
  10. Pre-Calc

    Does y=1/x have an inverse? It is a one-to-one function, so it should be the inverse equation is the same??? yes, the inverse is the same. Check it with G(f(x)) So, when drawing the inverse, it is just the same graph?

More Similar Questions