# Nature of Function (In Terms of a Graph)

For

*y*=*f*(*x*)

# Even Function

The graph of an even function will be symmetrical to Y-axis. It simply means that the graph on the left side of the Y-axis will be the same as the graph on the right side of the Y-axis.*y*=

*f*(

*x*) =

*x*

^{2}, the graph on the left hand side of the Y-axis is the same in shape and size with respect to the graph on the right hand side of the Y-axis.

**Properties of the graph of an even function**

- Even function graphs are one-fold graphs. i.e., we can make both the parts of the graphs overlap on each other, by just folding the graph paper once.
- Whenever we fold the paper, we fold it along the axis, either on the X-axis or on the Y-axis.

# Odd Function

For*y*=

*f*(

*x*)

The graph of an odd function will be symmetrical to the origin. It simply means that the graph on one side of the origin will be same as the graph on the other side of the origin.

As we have seen in the case of

As we have seen in the case of

*y*=*f*(*x*) = 1/*x*, the graph on one side of the origin is the same in shape and size with respect to the graph on the other side of the origin.(i) y = 1/x (ii) y = 1/x^{3} |

**Properties of the graph of an odd function**

- Odd function graphs are two-fold graphs, i.e., we can make both the parts of the graph overlap on each other, by folding the graph paper twice, once along the X-axis and then along the Y-axis.
- Whenever we fold the paper to overlap the graph, we fold it along the axis, either on the X-axis or on the Y-axis.

# Composite Function

Composite function is the kind of function which is composed into one by two or more than two functions.*y*=

*f*(

*x*) = 5

*x*+ 3 and

*y*=

*g*(

*x*) =

*x*

^{2}is given, then

*f*(

*g*(

*x*)) or

*g*(

*f*(

*x*)) is the composite function.

In the above example,

*f*(

*g*(

*x*)) = 5

*g*(

*x*) + 3 = 5

*x*

^{2}+ 3

*g*(

*f*(

*x*)) = [

*f*(

*x*)]

^{2}= (5

*x*+ 3)

^{2}

Example-1

If

*g*(*x*) = and*f*(*x*) =*x, x*â‰ 0 is given. Which of the following is true?*f(f(f(g(g(f(x))))) = g(g(f(g(f(x)))))**f(f(g(g(g(f(x))))) = g(g(f(g(f(x)))))**f(g(f(g(g(f(g(x)))))) = g(g(f(g(f(x)))))**f(f(f(g(g(f(x))))) = f(f(f(g(f(x)))))*

Solution

*f*(

*x*) =

*x*and

*g*(

*x*) = 1/

*x*

*f*â€™ any number of times is not changing the final result, but applying â€˜

*g*â€™ is making the value reciprocal. So, LHS can be equal to RHS only if the number of â€˜

*g*â€™ on both the sides is either even or odd.

So, option (b) is the answer.

# Inverse Function

If two functions
For example,

There lies an important relationship between function and graphs. We will see more application of this in the Chapter: Graphs and Maxima Minima.

*f*and*g*satisfy*g*(*f*(*x*)) =*x*for every*x*in the domain of f and similarly*f*(*g*((*x*)) =*x*for every*x*in the domain of*g*, then f is said to be the inverse of g and vice-versa. It is written as*f*^{-}^{1}and*g*^{-}^{1}.*y*= log_{e}*x*and*y*=*e**are inverse to each other.*^{x}Example-2

*y*=

*f*(

*x*) = . Find

*f*

^{-}âˆ’

^{1}(

*x*).

Solution

*y*= , or,

*y*â€“ 1 = âˆ’ 1 =

Or,

*x*+ 2 = , or,*x*=*f*

^{-}

^{1}(

*x*) =

There lies an important relationship between function and graphs. We will see more application of this in the Chapter: Graphs and Maxima Minima.

Example-3

A polynomial

*f*(*x*) with a real coefficient satisfies the functional equation*f*(*x*).*f**= f*(*x*) +*f*. If*f*(2) = 9, then*f*(4) is:- 82
- 17
- 65
- None of these

Solution

Assume

*f*(*x*) =*x**+ 1 and*^{n}So,

*f*(*x*).Now

*f*(*x*) =*x**+ 1 = 9*^{n}Or,

*x**= 8, or,*^{n}*n*= 3*f*(

*x*) =

*x*

^{3}+ 1

Hence,

*f*(*x*) = 4^{3}+ 1 = 65Example-4

Let

*f*(*x*) be a function such that*f*(*x*+ 1) +*f*(*x*âˆ’ 1) =*f*(*x*) for every real*x*. Then for what value of*p*is the relation*f*(*x*+*p*) =*f*(*x*) necessarily true for every real*x*?- 3
- 4
- 5
- 6

Solution

*f*(

*x*+ 1) +

*f*(

*x*âˆ’ 1) =

*f*(

*x*) â€¦1

Putting (

*x*+ 1) at the place of*x*or,

*f*(*x*+ 2) +*f*(*x*) =*f*(*x*+ 1) â€¦2Adding 1 and 2, we get

*f*(*x*+ 2) +*f*(*x*âˆ’ 1) = 0Similarly,

*f*(*x*+ 3) +*f*(*x*) = 0Or,

*f*(*x*+ 4) +*f*(*x*+ 1) = 0Or,

*f*(*x*+ 5) +*f*(*x*+ 2) = 0Or,

*f*(*x*+ 6) +*f*(*x*+ 3) = 0Now replacing

*f*(*x*+ 3) by â€“*f*(*x*),*f*(*x*+ 6) +*f*(*x*+ 3) =*f*(*x*+ 6) âˆ’*f*(*x*) = 0So,

*p*= 6