Informative line

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