Sketching inverse trig graphs (1 Viewer)

[eyeseeyou]

How would I sketch

y=2+(1/2)tan^-1(x-3)

Thanks

 

[InteGrand]

How would I sketch

y=2+(1/2)tan^-1(x-3)

Thanks

 

[eyeseeyou]

Yeah

Rules are .

domain: all x
Range: -pi/2<tan^-1(x-3)<pi/2 (please correct me on this if I am wrong)
Point of inflection: let x-3=0
x=3
Sub in LHS point, x=idk

Inverse tan curve (aka "mother curve") is this:
http://mathworld.wolfram.com/images/eps-gif/ArcTan_702.gif

 

[InteGrand]

 

[eyeseeyou]

Nah I have trouble with the range and "subbing in LHS, hence I said x=idk)

 

[eyeseeyou]

If you have something like -3<sin^-1 (x/4)<3 how would you find x?

 

[InteGrand]

Here are the required steps (in order):

1) Take the arctan graph and shift it right 3 units.

2) Compress this vertically by a factor of 2.

3) Shift this up 2 units.

 

[eyeseeyou]

Here are the required steps (in order):

1) Take the arctan graph and shift it right 3 units.

2) Compress this vertically by a factor of 2.

3) Shift this up 2 units.

Become wider or thinner?

 

[InteGrand]

Become wider or thinner?

"Thinner"

 

[Paradoxica]

How would I sketch

y=2+(1/2)tan^-1(x-3)

Thanks

By inspection.

 

[eyeseeyou]

Here are the required steps (in order):

1) Take the arctan graph and shift it right 3 units.

2) Compress this vertically by a factor of 2.

3) Shift this up 2 units.

So the origin for this graph is (3,2)?

 

[InteGrand]

So the origin for this graph is (3,2)?

The point on the graph of y = arctan(x) that was at (0,0) gets sent to (3,2) by the transformations, yes.

 

[eyeseeyou]

The point on the graph of y = arctan(x) that was at (0,0) gets sent to (3,2) by the transformations, yes.

Then what are the asymptotes?

 

[Dettol]

y=pi/4+2 and y=-pi/4+2

 

[eyeseeyou]

Help

1. Let f(x)=sin^-1 x
a. State the domain and rage of f(x)
b. Find f'(x) and f''(x)
c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points
d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0)
e. Sketch y=f(x) and y=x
f. For what value of m does the line y=mx cut y=f(x) at three points
g. Investigate the concavity of the curve and find the coordinates of the point of inflextion

 

[InteGrand]

Help

1. Let f(x)=sin^-1 x
a. State the domain and rage of f(x)
b. Find f'(x) and f''(x)
c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points
d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0)
e. Sketch y=f(x) and y=x
f. For what value of m does the line y=mx cut y=f(x) at three points
g. Investigate the concavity of the curve and find the coordinates of the point of inflextion

Which parts do you need help with? Do you know how to differentiate/graph/etc. the inverse sine function?

 

[InteGrand]

The whole thing

a. For domain I got -pi/2=<x=<pi/2
b. f'(x)=(sin^-1 x)' which idk

c-g idk either

You may want to check your textbook, lots of these are very basic and there's not much point doing this topic if you haven't learnt the basics yet.

 

[Trebla]

Help

1. Let f(x)=sin^-1 x
a. State the domain and rage of f(x)
b. Find f'(x) and f''(x)
c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points
d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0)
e. Sketch y=f(x) and y=x
f. For what value of m does the line y=mx cut y=f(x) at three points
g. Investigate the concavity of the curve and find the coordinates of the point of inflextion

Do you know what an inverse sine function is? If you are studying ahead (which I would advise against doing if you haven't yet mastered the basics of Year 11), I suggest you read how it is derived first.

 

[eyeseeyou]

Do you know what an inverse sine function is? If you are studying ahead (which I would advise against doing if you haven't yet mastered the basics of Year 11), I suggest you read how it is derived first.

Yes basically it's a sine curve which is restricted so that it becomes a one to one function

I plan (at the end of the year) to get up to binomial theorem and in the summer holidays, start on past papers

 

[aoc]

1a) domain is -1 to 1, range is -pi/2 to pi/2
b) first derivative is 1/sqaure root (1-x^2), second derivative shouldnt be too hard to do
c)it is defined at 1 and -1, but its derivative is not defined at x=1 or -1