INVERSE FUNCTIONS – Lesson #1

1.a) y=√4−x2

Inverse function of given equation is : y=√4−x2

Since quantity under root should be positive

Therefore, 4−x2>0

=> 4 ≥x2

=>x2≤4

=> −2≤x ≤2

Positive Domain : 0≤ x≤2

b. y=x2−6x+9

Inverse function of given quation is : y=√x+3

=> x>0

Positive Domain : 0≤ x≤∞

2. y=x+11−2 x

=>y (1−2 x )=x+1

=>y−2xy=x+1

=>y−1=x+2xy

=>x=y−11+2 y

Replacing x with y

So,y=x−11+2 x is the inverse of the given function.

b. y= e2x−1e2x+1

Applying componendo and dividendo:

y+1y−1

= e2x−1+e2x+1e2 x−1−e2x−1

y+1y−1

=2e2x

−2

y+1y−1

=−e2x

1

y+1y−1

=−e2x

1

x=12( log ( y+1 )−log (1− y ))

Replacing x with y

y=12(log ( x+1 )−log (1−x ))

c. y=x2+2 x+2

y=x2+2 x+1+1

y=¿

y−1=¿

√ y−1= x+1

x=√ y−1−1

Replacing x with y

y=√x−1−1

3. Inverse function of y = loge(2x+ 1)

e y=2 x+1

e y−1=2x

e y−12

=x

Replacing x with y

y= ex−12

Domain of the function : −∞≤x ≤∞

Range of the function : at x = −∞ y = -0.5

at x = 0 y = 0

at x = ∞ y = ∞

So Range (−0.5,∞)

4.

a) f(x) = ex

1+ex

Since x can take any value without violating any math rules:

So, Domain −∞≤x ≤∞

b) f(x) = ex

1+ex

DIfferentiting with respect to x

f’(x) = ddxex∗1+ex− d

dx1+ex∗ex

¿¿

= ex∗1+e x−ex∗e x

¿¿

= ex+e2x−e2x

¿¿

= ex

¿¿

DIFFERENTIATION USING INVERSE FUNCTIONS

1. dydx

= y2

Let Suppose , y=13−x

3−x=1y

x=−1y

+3

Differentiating with respect to y

dxdy

=−ddy

( 1y)+ ddy

(3)

dxdy

= 1

y2+0

dydx×dxdy

= y2∗1y2

dydx×dxdy

=1

Hence Proved

2.dydx

= y−3

Let Suppose , x= y4

4+1

DIfferentiting with respect to y

dxdy

= ddy

( y4

4)+ ddy

(1)

dxdy

= y3+0

dydx×dxdy

= y−3∗y3

¿1

Hence Proved

3. dydx

=2 y+3

Let Suppose , 18 y=e2x−27

18 y+27=e2x

2 x=ln (18 y+27)

DIfferentiting with respect to y

2dxdy

= ddyln (18 y+27)

2dxdy

= 1818 y+27

dxdy

= 918 y+27

dxdy

= 12 y+3

dydx×dxdy

=2 y+ 3∗12 y+3

dydx×dxdy

=1

Hence Proved

Page 4

Examples:1. Evaluate the following:

a. cos−1 ¿ = π /3

b. cos−1(−12

¿)¿=2π /3

c. sin−1( 1√2

)=π /4

d. sin−1(−√32

¿)¿=4 π /3

e. tan−1√3=π /3

f. tan−1(−1√3

)=5 π6

∧11 π

6

2. Evaluate the following:

a. cos−1(cosπ6) =

π6∧5 π

6

b. cos (sin−1 1

√2) = 1/√2

c. sin(cos−1(−√32

))=1/2

d. sin−1 12+cos−1 1

2=π /2

3.

a. Consider the function f(x) = 2cos−1x2

i. State the domain and range of f(x)Solution :Domain : −2≤x ≤2

Range:- 0≤ f (x )≤2π

ii. Draw the graph of f(x)

b. State the domain and range of y=3sin−1(x−2) and sketch its graphSolution :Domain : 1≤x ≤3

Range:- −3π2≤ f (x )≤ 3 π

2

Graph:-

c. State the range of y=3 tan−12x

Solution :Domain : −∞≤x ≤∞

Range:- −π2≤ f (x )≤ π

2

d. State the domain and range of y=x sin−1 x

Solution :Domain : −1≤x ≤1

Range:- 0≤ f (x )≤π2

Page 5:

Proving some of these results – this is examinable!1. sin−1¿

Let sin−1¿ i.e sin y=−x so that x=−sin y,i.e x=sin (− y ) Hence sin−1¿

Therefore sin−1¿

2. cos−1(−x )=π−cos−1 x

Let cos−1 ¿ i.e cos y=−x so that x=−cos y=cos (π− y),i.e x=cos (− y ) Hence cos−1 ¿

Therefore cos−1 ¿

3. tan−1 ¿

Let tan−1 ¿ i.e tan y=−x so that x=−tan y,i.e x=tan(− y ) Hence tan−1 ¿

Therefore tan−1 ¿

4. sin−1 x+cos−1 x= π2

Let sin−1¿ then x=sin y=cos(π2− y )

Therefore cos−1 ¿

Examples Page 6:

1. Show that cos−1513

=sin−1 1213

Suppose cos−1513

=x

Then cos x=513

And sin2 x+cos2 x=1

Therefore sin x=1213

x=sin−1 1213

Hence,cos−1513

=sin−1 1213

2. Show that tan−134=sin−1 3

5

Suppose tan−134=x

Then tan x=34

Since tan x=Prependicualar

Base

In Right Angled TriangleBase2+Prependicular2=Hypotenuse2

Hypotenuse2=32+52

Hypotenuse=√25=5

sin x= PrependicualarHypotenuse

=35

Thus,

x=sin−1 34

Hence ,tan−134=sin−1 3

5

3. cos (sin−1 23)

Suppose,sin−1 23=x

sin x=23

So, Equation becomes cos xAnd sin2 x+cos2 x=1

Therefore cos x=√53

4. cos¿

Suppose,cos−123=x

cos x=23

So, Equation becomes cos2 xAnd cos2 x=2cos2 x−1

=> cos2 x=2( 22

32)−1

=>cos2 x=89−1

=>cos2 x=−19

5.a. cos¿Suppose,sin−1 x=t

Therefore,x=sin tSo, Equation becomes cos tAnd sin2 t+cos2 t=1=>cos2t=1−x2

=>cos t=√1−x2

b.sin ¿

Suppose,cos−1t4=x

Therefore,t4=cos x

So, Equation becomes sin xAnd sin2 x+cos2 x=1

sin2 x=1− t2

42❑

sin x=√1− t2

16

c.tan¿Suppose,tan−1 x=tTherefore,x=tan tSo, Equation becomes tan2 t

And tan2 t=2 tant

1−tan2t

=>tan2 t=2 x

1−x2

d.sin ¿

Suppose,tan−1√x2−2 x=t=>√ x2−2x=tan tSquaring both sides, we have:x2−2 x=tan 2tAdding 1 to both sides=> (x−1)2=tan2t+1=>(x−1)2=sec2t

=>x−1=sec t

=>cos t=1x−1

So, Equation becomes sin 2 tsin 2 t=2∗sin t∗costAnd sin2 t+cos2 t=1=>sin2 t=1−cos2t

=>sin t=√1− 1(x−1)2

=>sin

2 t=2∗√x2−2 xx−1

∗1

x−1

=> sin 2 t=2√ x2−2x

(x−1)2

6.

sin−1 x+cos−1 y= π12 (I)

sin−1 y−cos−1 x=7π12

(II)

Suppose,sin−1 x=a, so cos−1 x=π2−a

Suppose,sin−1 y=b, so cos−1 y=π2−b

Above equation becomes:

π2−b+a= π

12

b−π2

+a=7 π12

Adding both sides:

2a=8 π12

a=π3

=> b=π4

sin−1 x=π3=¿ x=sin π

3

sin−1 y= π4=¿ y=sin π

4