inverse
posted by Jen .
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.
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