A Function can be classified as Even, Odd or Neither. This classification can be determined graphically or algebraically.
How to check if the graph is odd ?
The graph will be symmetric with respect to the origin.
In other words :
If you spin the picture upside down about the Origin, the graph looks the same!
How to check if the graph is even ?
The graph will be symmetric with respect to the yaxis.
Properties of odd function :
Properties of even function :
Problem 1 :
Which function is an even function?
a. y = h(x) = √x b. y = p(x) = x^{3}
c. y = R(x) = 1/x d. y = x
Solution :
Option a :
y = h(x) = √x
Put x = x
h(x) = √x
The function is not defined for negative values
Option b :
y = p(x) = x^{3}
Put x = x
p(x) = (x)^{3}
p(x) = x^{3}
It is odd function.
Option c :
y = R(x) = 1/x
Put x = x
R(x) = 1/(x) ==> 1/x
R(x) = 1/x
It is odd function.
Option d :
y = x
Put x = x
y = x
y = x
that is, f(x) = f(x). It is even function.
Problem 2 :
Examine these functions. Which statement is correct?
a. Graph I is the graph of an even function.
b. Graph II is symmetric to the xaxis.
c. Graph III is symmetric to the yaxis.
d. Graph IV is an odd function.
Solution :
Option a :
Graph I is the graph of an even function.
Option b :
Graph II is symmetric to the xaxis.
Here the yaxis is like a mirror.
Option c :
It is symmetric about origin. But option c says, it is symmetric to the yaxis. So, it is incorrect.
Option d :
It is symmetric about origin. So, option d is correct.
Problem 3 :
Which function could be an even function?
a) A function known as d(x) where d(3) = 10 and d (3) = 10
b) A function known as k(x) where k(7) = 20 and k(7) = 20
c) A function known as M(x) where M(4) = 10 and M (4) = 0
d) A function known as C(x) where C(7) =  6 and C(7) = 6
Solution :
Even function will be symmetric about yaxis.
All points that lie on the curve will be on the rule given below,
(x, y) ==> (x, y)
Option a :
d(3) = 10 and d (3) = 10
When x = 3, y = 10
when x = 3, y = 10
It does not satisfy the rule above. So, it is not symmetric about yaxis and it is not even.
Option b :
k(7) = 20 and k(7) = 20
When x = 7, y = 20
when x = 7, y = 20
It satisfies the above rule. So, it is even function.
Problem 4 :
Start with the incomplete graph of p(x) as shown. Complete the graph if p(x) is symmetric to the origin
Solution :
Even function will be symmetric about origin.
All points lie on the curve will follow the rule given below.
(x, y) ==> (x, y)
Before the rule

After the rule

Problem 5 :
Start with the incomplete graph of f(x) as shown. Complete the graph if f(x) is symmetric to the yaxis.
Solution :
Even function will be symmetric about yaxis.
All points lie on the curve will follow the rule given below.
(x, y) ==> (x, y)


Problem 6 :
If g(x) is an odd function and g(5) = 7, which one of the following must be true?
a. g(5) = 7 b. g(5) = 7 c. g(7) = 5 d. g(7) =  5
Solution :
Odd function is symmetric about origin. The points that lie on the curve will follow the rule below.
(x, y) ==> (x, y)
If g(5) = 7, the g(5) = 7
Option a is correct.
Problem 8 :
Which function is not even (symmetric to the yaxis)?
a. y = 4 x^{2} + 3 b. y = 2 x  6
c. y = x^{2 }+6x + 5 d. y = 6
Solution :
If the function satisfies the condition, f(x) = f(x), then it is even function.
Option a :
f(x) = 4 x^{2} + 3
f(x) = 4 (x)^{2} + 3
f(x) = 4 x^{2} + 3
f(x) = f(x)
It is even
Option b :
f(x) = 2 x  6
Put x = x
f(x) = 2x  6
f(x) = 2x  6
f(x) = f(x)
So, it is even.
Option c :
f(x) = x^{2 }+6x + 5
f(x) = (x)^{2 }+ 6(x) + 5
f(x) = x^{2 } 6x + 5
f(x) ≠ f(x)
It is not even. So, option c is correct.
May 21, 24 08:51 PM
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