Inverse functions and tangent lines

In summary: I don't understand what you mean by "odd multiples of π")No, this is incorrect. The line y = x - 2, is tangent to y = f(x) at all points where x = 2πn, so you luck out.
  • #1
wuffle
26
0

Homework Statement


2 problems, i solved both of them but I am not 100 % I am right

Find all points on the curve y=x-2cosx where the tangent line to the curve is parallel to the line y=x and write an equation of the tangent line at such point

Let f^-1 be the inverse of the one -to- one function f(x)=x^3+x.

Find (f^-1)'(10).



Homework Equations


slope of tangent line=f'(of the point)


The Attempt at a Solution




1st.

This seemed tricky and I did not figure it out till I had a good 11 hour sleep, what I did:

y=x , parallel lines have same slopes=>slope of tangent line is 1

y'=1+2sinx

1+2sinx=1

x=0+2pi*n

we found all points on the curve where tangent is parallel to y=x

lets find y coordinate
y=0-2cos(0)=-2

equation of tangent line
y+2=1(x-0)
y=x-2

2nd.

I think this has a twist in it but I can't find it :( , here's what I did

f^-1(x)=1/(f'(f^-1(x))

we need to find f^-1(10)
since we got f(x) its not hard to do so
10=x^3+x
x=2

f^-1(x)=1/(f'(2))
f'=3x^2+1
f^-1(x)=1/13

this is the answer?

Please check what I have written, I apologize for such a mess t.t
 
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  • #2
Problem #1:
You are missing some roots for the equation, [itex]\sin(x) = 0\,.[/itex] Look at the unit circle. For what angle(s) is y = 0 ?

Because you are missing some of the solutions, you are also missing at least one tangent line. Also, why is it that a single tangent line is sufficient for all the values of x that you did find?
 
  • #3
wuffle said:

Homework Statement


2 problems, i solved both of them but I am not 100 % I am right

Find all points on the curve y=x-2cosx where the tangent line to the curve is parallel to the line y=x and write an equation of the tangent line at such point

Let f^-1 be the inverse of the one -to- one function f(x)=x^3+x.

Find (f^-1)'(10).



Homework Equations


slope of tangent line=f'(of the point)


The Attempt at a Solution




1st.

This seemed tricky and I did not figure it out till I had a good 11 hour sleep, what I did:

y=x , parallel lines have same slopes=>slope of tangent line is 1

y'=1+2sinx

1+2sinx=1

x=0+2pi*n
Actually sin(x) = 0 for x = n*pi, not just multiples of 2pi.
wuffle said:
we found all points on the curve where tangent is parallel to y=x

lets find y coordinate
y=0-2cos(0)=-2
No. This is just one of the points at which the tangent line has a slope of 1. Your problem statement says "write an equation of the tangent line at such point". It's not clear to me what this means.
wuffle said:
equation of tangent line
y+2=1(x-0)
y=x-2

2nd.

I think this has a twist in it but I can't find it :( , here's what I did

f^-1(x)=1/(f'(f^-1(x))

we need to find f^-1(10)
since we got f(x) its not hard to do so
10=x^3+x
x=2

f^-1(x)=1/(f'(2))
f'=3x^2+1
f^-1(x)=1/13

this is the answer?

Please check what I have written, I apologize for such a mess t.t
 
  • #4
Problem #2:

What you have done is correct.

For exponents and/or superscripts, use the Advanced message screen. (Click on "Go Advanced" below.)

Once on the advanced message screen, use the "X2" button above the message box.

Then you can write (f^-1)' (x)=1/(f '(f^-1(x)) as (f -1)' (x)=1/(f  '(f -1(x))
 
  • #5
SammyS said:
Problem #1:
You are missing some roots for the equation, [itex]\sin(x) = 0\,.[/itex] Look at the unit circle. For what angle(s) is y = 0 ?

Because you are missing some of the solutions, you are also missing at least one tangent line. Also, why is it that a single tangent line is sufficient for all the values of x that you did find?

Yep, you are right, sinx=0, x=pi n , not 2pi n.

sinx=0

x=pi * n

for x=pi

y=pi+2

equation of tangent line

y-(pi+2)=1*(x-pi)

Well, it asks to write an equation of the tangent line, so I was thinking it was only 1, guess I was wrong.But can't I write infinitive amount of tangent lines?since of n?what do I do then?Do I also write for n =-1 ,+-2, +-3?
 
Last edited:
  • #6
Prob. #1:

cos(±π) = -1

= cos(±3π) = cos(±5π) = cos(±7π) = ...

Also, the line y = x - 2, is tangent to y = f(x) at all points where x = 2πn, so you luck out. (You probably should show that this is the case.)

A different line is tangent to y = f(x) for x = odd multiples of π .
 
  • #7
SammyS said:
Prob. #1:

cos(±π) = -1

= cos(±3π) = cos(±5π) = cos(±7π) = ...

Also, the line y = x - 2, is tangent to y = f(x) at all points where x = 2πn, so you luck out. (You probably should show that this is the case.)

A different line is tangent to y = f(x) for x = odd multiples of π .

Thanks a lot for your help!
 

1. What is an inverse function?

An inverse function is a function that "undoes" the action of another function. If a function f(x) takes an input x and produces an output y, its inverse function f^-1(y) takes an input y and produces an output x.

2. How do I find the inverse of a function?

To find the inverse of a function, you can follow these steps:
1. Replace f(x) with y.
2. Swap the positions of x and y.
3. Solve for y in terms of x.
4. Replace y with f^-1(x).
The resulting expression is the inverse function.

3. How are inverse functions related to tangent lines?

The inverse of a function and the tangent line to that function at a specific point are related in that the slope of the tangent line at that point is equal to the reciprocal of the slope of the inverse function at the corresponding point.

4. What is the derivative of an inverse function?

The derivative of an inverse function can be found by using the formula:
(f^-1(x))' = 1 / (f'(f^-1(x))).
In words, this means that the derivative of an inverse function is equal to the reciprocal of the derivative of the original function evaluated at the inverse function's input.

5. How do I use inverse functions and tangent lines in real-world applications?

Inverse functions and tangent lines are used in many real-world applications, such as in physics, engineering, and economics. For example, inverse functions can be used to model the relationship between two variables, and tangent lines can be used to approximate the behavior of a function at a specific point. These tools can also be used to optimize processes, solve equations, and analyze data.

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