3u Questions - Inverse (1 Viewer)

[gh0stface]

Hi, having trouble with a few questions regarding the inverse topic.

1. State the domain and range of each of the following functions:

a) y=InverseTan (Sqrt (1-x^2))


b) y=InverseCos (Sqrt (1/4-x^2))


c) y= xInverseSin (x^2)


I could get the domain, however the range is weird, answers for range are

a) 0<=y<=Pi/4, b) Pi/3<=y<=Pi/2, c) -Pi/2<=y<=Pi/2

Someone explain how to get it?





2. Sketch the graph of the deriviative of y=InverseSin(x), y=InverseCos(x), y=InverseTan(x),

Apparently for inverse sin its a concave up parabola with vertex 0,1, the vertex while inverse cos in concave down with vertex 0,-1. Tan is some weird hill-like curve with symmetry about the y axis. (WOOPS sorry for the mistake)




3. Find the inverse of the curve, also stating its domain and range.

y= - Sqrt(9-16x^) 0<=x<=3/4

Isn't this curve a semi circle? the answer is like half a parabola.


4. Find the value of InverseSin (1/Sqrt(2))

For this question if theres no restriction given, then wat would be an appropriate answer? pi/4 or 3pi/4

Does it involve using general solution of sin to find out? that is

x = (-1)^k . InverseSin(a) + kpi


Im really confused about the purpose of the general solutions of sin,cos,tan? when do you apply it?


I heard when doing questions like these, the angles can only be in the 1st or 4th quadrant for sin and tan, and 1st and 2nd for cos. Why is it this?


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I know this is alot of questions, but im really stuck, any help would be greatly appreciated.


thanks

 

[namburger]

2. Sketch the graph of the deriviative of y=InverseSin(x), y=InverseCos(x), y=InverseTan(x)

Inverse sin(x)
Derivative is y = 1/ root (1-x^2)
The answer seems wrong, as the point 0,0 doesn't satisfy the above equation.
If you diffrentiate it to find stationary pts, you will get x/ (1-x^2)^(3/2)
let y' = 0, x = 0 --> sub it into y and the stationary pt occurs at 0,1
The parabola would be a concave up parabola with a vertex , (0,1) between the domain -1 < x < 1

Inverse cos(x)
Derivative is y = -1/root (1-x^2)
y' = -x/ (1-x^2)^(3/2)
let y' = 0, x = 0 --> stationary pt occurs at 0,-1
Parabola would be concave down with vertext, (0,-1) between -1 < x < 1

Inverse tan(x)
Derivative is y = 1/root (1+x^2)
The domain of this curve is all values. to find the horizontal asymptote, lim x-> infinite and -infinite. horizontal asymptote occurs at y=0
Find stationary pts:
y' = -x / (1+x^2)^(3/2)
let y' = 0, and a stationary pt occurs at 0,1

3. Find the inverse of the curve, also stating its domain and range.

y= - Sqrt(9-16x^) 0<=x<=3/4

Isn't this curve a semi circle? the answer is like half a parabola.

- root (9-16x^2) = - root (16(9/16-x^2))
= - 4 root (9/16-x^2)
The 4 in front makes it not a semi-circle

find y' and there will be a stationary pt at (0,-3). sub x=3/4 into y and the range is from -3 < y <0

4. Find the value of InverseSin (1/Sqrt(2))

For this question if theres no restriction given, then wat would be an appropriate answer? pi/4 or 3pi/4

by definition for the equation y = sininverse (x), the range of inverse sin is between -pi/2 < y < pi/2
So the answer is pi/4

Im really confused about the purpose of the general solutions of sin,cos,tan? when do you apply it?

You could use the general solution, but remember what the ranges are. Only choose the answers to the general solutions that are in the range

I heard when doing questions like these, the angles can only be in the 1st or 4th quadrant for sin and tan, and 1st and 2nd for cos. Why is it this?

y = Inversesin (x) Range: -pi/2 < y < pi/2
Only in 1st and 4th
y = InverseTan (x) Range: -pi/2 < y < pi/2
Only in 1st and 4th
y = inversecos (x) Range: 0 < y < pi
Only in 1st and 2nd

 

[YannY]

neat stuff nam lets own wow together

 

[namburger]

neat stuff nam lets own wow together

After HSC lol, i cant do question 1 =[

 

[watatank]

Hi, having trouble with a few questions regarding the inverse topic.

1. State the domain and range of each of the following functions:

a) y=InverseTan (Sqrt (1-x^2))


b) y=InverseCos (Sqrt (1/4-x^2))


c) y= xInverseSin (x^2)


I could get the domain, however the range is weird, answers for range are

a) 0<=y<=Pi/4, b) Pi/3<=y<=Pi/2, c) -Pi/2<=y<=Pi/2

Someone explain how to get it?

find the domain of the inverse function of y and that's your range of the normal y function.

 

[namburger]

find the domain of the inverse function of y and that's your range of the normal y function.

Ok i don't think this works: for example
a) y=InverseTan (Sqrt (1-x^2))
x = InverseTan (Sqrt (1-y^2))
Tan x = Sqrt (1-y^2)
(Tan x)^2 = 1-y^2
y^2 = 1 - (Tan x)^2
1 - (Tan x)^2 >= 0
1 >= (Tan x)^2
Therefore domain: -pi/4 < x < pi/4
Not same as answer

 

[YannY]

y=InverseTan (Sqrt (1-x^2)) The domain of inversetan is boundless hence the domain of sqrt(1-x^2) must not be negative. i.e 1-x^2>0 i.e x^2<1 therefore -1<x<1.

Edit: highest point is when x=0 and lowest point is when x=-1,x=1 therefore range is 0<=y<=Pi/4

There ya go wow buddy nam =]