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### Inverse Function And Its Derivative

Learn how to find the inverse function of a one to one function ?f?. Solve inverse of a function examples and horizontal line test.

# One-One Functions

• A function $$'f'$$ is called one to one function if it never takes some value twice, that is

$$f(x_1)\neq f(x_2)$$,  whenever $$x_1\neq x_2$$

• ''Consider a function mapped from Set A to Set B''
• One to one function

• Not a one to one function, since, $$f(2)=f(3)=5$$

• A function is one to one if no value in set B has more than one pre image in set A.

#### Some functions $$'f'$$ are given by a table as shown below, pick the function which is one to one.

A x 1 2 3 4 5 6 7 f(x) 5.2 6.8 1.2 2.7 5.2 6.8 0

B x 1 2 3 4 5 6    7 f(x) 2.8 3.2 1.9 3.2 1.1 7 18

C x 1 2    3 4 5 6 7 f(x) 1.8 2 3.4 5.6 2.1 1.9 11

D x 1 2 3 4 5    6    7 f(x) 4.6 2.8 9.2 2.8 1 5 3.2

×

Observe the table, one-one function is the one, in which no value is repeated in the row of values of $$f(x)$$.

In option 'a' $$\to f(1)=f(5)\to$$ Not one to one

In option 'b' $$\to f(2)=f(4)\to$$ Not one to one

In option 'd' $$\to f(2)=f(4)\to$$ Not one to one

$$\therefore$$ option 'c' is correct.

### Some functions $$'f'$$ are given by a table as shown below, pick the function which is one to one.

A

 x f(x) 1 2 3 4 5 6 7 5.2 6.8 1.2 2.7 5.2 6.8 0
.

B

 x f(x) 1 2 3 4 5 6 7 2.8 3.2 1.9 3.2 1.1 7 18

C

 x f(x) 1 2 3 4 5 6 7 1.8 2 3.4 5.6 2.1 1.9 11

D

 x f(x) 1 2 3 4 5 6 7 4.6 2.8 9.2 2.8 1 5 3.2

Option C is Correct

# Horizontal Line Test

• A function is one to one if no horizontal line intersects its graph more than once.

• One of the easy ways to observe this to check, if there is a local maximum or local minimum point. If these points are present, we say function is not one to one, otherwise it is one to one.

#### Graph of some function are given below in the option. Choose the function which is one to one.

A

B

C

D

×

Using the horizontal line test, the curve must be cut at only one point for the function to be one to one.

Hence, option A is incorrect.

Hence, option B is incorrect.

Hence, Option C is correct.

Hence, Option D is incorrect.

### Graph of some function are given below in the option. Choose the function which is one to one.

A
B
C
D

Option C is Correct

# Graph of Inverse function[f-1(x)] with respect to Graph of f(x)

• The graph of $$f^{-1}$$ is obtained by reflecting the graph of $$f$$ about the line $$y=x$$.

• For every point $$(h,\,k)$$ on the graph of $$f$$ there will be a point $$(k,\,h)$$ on the graph of $$f^{-1}(x)$$. Since $$(h,\,k)$$ and $$(k,\,h)$$ are mirror images of each other in the line $$y=x$$, the entire graph of $$f$$ and $$f^{-1}$$ will be as shown.

#### Which of the following can be the graph of $$f^{-1}$$ if the graph of $$f$$ is

A

B

C

D

×

Take the reflection of the graph of $$f$$ about the line $$y=x$$.

Hence, correct option is (A).

### Which of the following can be the graph of $$f^{-1}$$ if the graph of $$f$$ is

A
B
C
D

Option A is Correct

# Finding the Value of Inverse Function from the Graph of Function 'f'

• If $$(h,\,k)$$ is on the graph of function $$f$$ then, $$f(h)=k$$ and $$f^{-1}(k)=h$$.

#### Given below is the graph of a certain function $$f(x)$$. what is the value of $$f^{-1}(-2)$$ ?

A 4

B 3

C –3

D 5

×

Draw a horizontal line $$y=-2$$ it will intersects the graph at $$(4,\,-2)$$.

As we know,

$$f(\alpha)=\beta$$ then $$f^{-1}(\beta)=\alpha$$

$$\therefore\,f(4)=-2$$

$$\Rightarrow\,f^{-1}(-2)=4$$

### Given below is the graph of a certain function $$f(x)$$. what is the value of $$f^{-1}(-2)$$ ?

A

4

.

B

3

C

–3

D

5

Option A is Correct

# Inverse of a Function

• If $$f$$ is a one to one function whose domain and range are A and B, then its inverse function denoted by $$f^{-1}$$ has domain B and range A  is defined by

$$f^{-1}(y)=x\iff f(x)=y$$

for any $$y$$ in $$B$$.

$$f(2)=4 \Rightarrow\,f^{-1}(4)=2$$

$$f(3)=9 \Rightarrow\,f^{-1}(9)=3$$

$$f(4)=6 \Rightarrow\,f^{-1}(6)=4$$

• Domain of $$f$$ = range of $$f^{-1}$$
• Range of $$f$$ = domain of $$f^{-1}$$
• $$f^{-1}$$ reverse the effect of $$'f'$$ and brings back $$x$$ to its original value.

• $$f^{-1}(f(x))=x$$

or

$$f(f^{-1}(x))=x$$

#### For a one to one function $$'f'$$ if $$f(2)=-7$$ then ,$$f^{-1}(-7)=$$

A 5

B 2

C –2

D 7

×

As we know,

$$f(\alpha)=\beta$$ then, $$f^{-1}(\beta)=\alpha$$

$$f^{-1}(-7)=2$$  because $$f(2)=-7$$

### For a one to one function $$'f'$$ if $$f(2)=-7$$ then ,$$f^{-1}(-7)=$$

A

5

.

B

2

C

–2

D

7

Option B is Correct

# Finding the Inverse Function of a One to One Function

1. Write $$y=f(x)$$
2. Express $$x$$ in terms of $$y$$ (if possible)
3. Interchange $$x$$ and $$y$$. The resultant $$y$$ is the required function $$f^{-1}(x)$$

e.g.

Step-1:  $$f(x)=2x^3+3$$

$$\Rightarrow\,y=2x^3+3$$

$$\Rightarrow x^3=\dfrac {y-3}{2}$$

Step-2:  $$\Rightarrow\,x=\left(\dfrac{y-3}{2}\right)^{1/3}$$

Step-3: $$y=\left(\dfrac{x-3}{2}\right)^{1/3}=f^{-1}(x)$$

#### Find a formula for inverse of the function $$f(x)=\dfrac{2x-3}{5x+1}$$

A $$f^{-1}(x)=\dfrac{x+3}{2-5x}$$

B $$f^{-1}(x)=\dfrac{5x+1}{2x-3}$$

C $$f^{-1}(x)=sinx$$

D $$f^{-1}(x)=\dfrac{x^2+1}{2x^2+3}$$

×

Step 1:

$$\to y=\dfrac{2x-3}{5x+1}$$

Step  2:

$$\to 5xy+y=2x-3$$

$$\Rightarrow\,x(5y-2)=-3-y$$

$$\Rightarrow\,x=\dfrac{-3-y}{5y-2}$$

Step 3:

$$\to y=\dfrac{-3-x}{5x-2}$$

$$\Rightarrow\,f^{-1}(x)=\dfrac{x+3}{2-5x}$$

### Find a formula for inverse of the function $$f(x)=\dfrac{2x-3}{5x+1}$$

A

$$f^{-1}(x)=\dfrac{x+3}{2-5x}$$

.

B

$$f^{-1}(x)=\dfrac{5x+1}{2x-3}$$

C

$$f^{-1}(x)=sinx$$

D

$$f^{-1}(x)=\dfrac{x^2+1}{2x^2+3}$$

Option A is Correct

# The Derivative of Inverse Function

• If $$f$$ is a one to one differentiable function with inverse function $$f^{-1}$$ then, $$f^{-1}$$ is also differentiable at $$x=a$$.

$$\left(f^{-1}\right)'(a)=\dfrac{1}{f'\left(f^{-1}(a)\right)}\:\:\:\:\:\:\:\rightarrow\,(1)$$

i.e. the derivative of $$f^{-1}$$ at $$x=a$$ is the reciprocal of the derivative of $$f$$ at $$f^{-1}(a)$$.

#### If $$f$$ is a one to one, differentiable function and $$f(7)=6$$ and $$f'$$$$(7)=2$$ then, find the value of $$\left(f^{-1}\right)'(6).$$

A $$\dfrac{1}{2}$$

B 7

C –6

D –7

×

$$\left(f^{-1}\right)'(a)=\dfrac{1}{f'\left(f^{-1}(a)\right)}$$

Put $$a=6$$,

$$\Rightarrow$$ $$\left(f^{-1}\right)'(6)=\dfrac{1}{f'\left(f^{-1}(6)\right)}$$

$$=\dfrac{1}{f'(7)}$$

$$=\dfrac{1}{2}$$

### If $$f$$ is a one to one, differentiable function and $$f(7)=6$$ and $$f'$$$$(7)=2$$ then, find the value of $$\left(f^{-1}\right)'(6).$$

A

$$\dfrac{1}{2}$$

.

B

7

C

–6

D

–7

Option A is Correct

# Finding the Derivative of Inverse Function at a Particular Value of x

Suppose we are given an expression for a function $$'f'$$, to the derivative of $$f^{-1}$$ at particular value either we need to find out the inverse function expression which is sometimes not possible or use the formula.

$$\left(f^{-1}\right)'(a)=\dfrac{1}{f'\left(f^{-1}(a)\right)}$$

• So find $$f^{-1}(a)$$ by observation and then, find $$f'$$ to get the result.

#### If $$f(x)=3x+2cos\,x+sin\,x$$ ,then the value of $$\left(f^{-1}\right)'(2) \,\,\:is$$

A $$\dfrac{1}{2}$$

B $$\dfrac{1}{4}$$

C 4

D 2

×

$$\left(f^{-1}\right)'(2)=\dfrac{1}{f'\left(f^{-1}(2)\right)}$$

Now observe that,

$$f(0)=3×0+2cos\,0+sin\,0$$

$$\Rightarrow\,f(0)=2$$

$$\Rightarrow\,f^{-1}(2)=0$$

$$\therefore\,\left(f^{-1}\right)'(2)=\dfrac{1}{f'(0)}$$

$$f'$$$$(x)=3-2sin\,x+cos\,x$$

$$\Rightarrow$$ $$f'$$$$(0)=3+1=4$$

$$\therefore\,\left(f^{-1}\right)'(2)=\dfrac{1}{4}$$

### If $$f(x)=3x+2cos\,x+sin\,x$$ ,then the value of $$\left(f^{-1}\right)'(2) \,\,\:is$$

A

$$\dfrac{1}{2}$$

.

B

$$\dfrac{1}{4}$$

C

4

D

2

Option B is Correct