**Functions**:

**Sketch of curve y = sec(sec**^{-1}x):

Domain

R - (-1, 1)

& range y = x

y

R - (-1, 1)

y = sec(sec

^{-1}x) = x, only when x

(-

, -1)

(1,

)

**Sketch of y = cot(cot**^{-1}x):

Domain

R - (-1, 1)

& range y = x

y = R

We should sketch

y = cot(cot

^{-1}x) = x

~~v~~ x

R

**Sketch of y = sin**^{-1}:

Domain [-1, 1]

~~v~~ x

R

& range y

b/c y = sin

^{-1}
y

**Dumb Question**: How domain is [-1, 1]

~~v~~ x

R ?

Ans:

i.e.

2|x|

1 + x

^{2} |x|

^{2} - 2|x| + 1

0

(|x| - 1)

^{2} 0

x

R

Let x = tan

y = sin

^{-1}(sin2

) =

This curve has sharp edge at x = ± 1

So, not differentiable at x = ± 1

**Dumb Question**: How does y =

- 2

for 2

>

Ans: Since our y

. So, if 2

>

to make it we substract it from

- 2

because sin(

- 2

) = sin2

**Sketch of y = cos**^{-1}
For domain

1

|1 - x

^{2}|

1 + x

^{2}
Which is true for all n; as 1 + x

^{2} > 1 - x

^{2} x

R

domain

[-1, 1]

For range

y =cos

^{-1}
y

(0,

)

Let x = tan

y = cos

^{-1} = cos

^{-1}(cos2

)

**Dumb Question**: How cos

^{-1}(cos2

) = - 2 tan

^{-1}x when x < 0 or

< 0

Ans: Range of y

[0,

]. So, when

< 0 or x < 0

Then we have to make it b/w [0,

]

So, cos

^{-1}(cos2

) = - 2

when

< 0

because

cos(- 2) = cos2 |

This curve has sharp edge at x = 0. So, not differentiable at x = 0.

**Sketch of y = tan**^{-1}
For domain

R except 1 - x

^{2} = 0

i.e. x

± 1

or x

R - {1, -1}

domain

R

Range y = tan

^{-1}
as y = tan

^{-1} y

Let x = tan

y = tan

^{-1} = tan

^{-1}(tan2

) =

**Dumb Question**: Why y = tan

^{-1}(tan2

) = -

+ 2 tan

^{-1}x, for

>

?

Ans: Range of y

& when

>

So, 2

>

.

To make it within range we substract

So, it comes under range.

Because of tan(-

+ 2

) = tan2

, this can be done.

This curve is neither continous nor differentiable at x = {-1, 1}

**Sketch of curve y = tan**^{-1}
For domain,

y = tan

^{-1}
x

r except 1 - 3x

^{2} = 0

x

±

x

R - {±

}

domain

R

For range

y = tan

^{-1}
y

Let x = tan

y = tan

^{-1} = tan

^{-1}(tan3

) =

**Dumb Question**: Why tan

^{-1}(tan3

) =

+ 3

if

< -

?

Ans: Range of tan

^{-1}(tan3

) is

but if

< -

So, 3

< -

So, to make it within given range. We add

to it.

and in III

^{rd} quadrant.

tan(

+ 3

) = tan3

So, this is done.

This curve is neither cont. nor differentiable at x = ±

.

**Sketch of y = sin**^{-1}(3x - 4x^{3}):

For domain

y = sin

^{-1}(3x - 4x

^{3})

x

[-1, 1]

For range

y = sin

^{-1}(3x - 4x

^{3})

y

Let x = sin

y = sin

^{-1}(sin3

) =

This curve has sharp edge at x = ± 1/2 so, not differentiable at x = ± 1/2.