Inverse functions and tangent lines

[wuffle]

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

 

[SammyS]

Problem #1:
You are missing some roots for the equation, $\sin(x) = 0\,.$ 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?

 

[Mark44]

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.

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.

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

 

[SammyS]

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))

 

[wuffle]

Problem #1:
You are missing some roots for the equation, $\sin(x) = 0\,.$ 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?

 

[SammyS]

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 .

 

[wuffle]

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!