Problem 1: A function f : R → R f : R → R is given by :

f x = x 3 f x = x 3

Is the function one-one.

Solution :

Statement of the problem : Draw a line parallel to x-axis to intersect the plot of the function as many times as possible.

Figure 1: The line parallel to x-axis intersects function plot at one point.

Function type

We find that all lines drawn parallel to x-axis intersect the plot only once. Hence, the function is one-one.

Problem 2: A function f : R → R f : R → R is given by :

f x = x 2 f x = x 2

Is the function one-one.

Solution :

Statement of the problem : We can solve f x 1 = f x 2 f x 1 = f x 2 and see whether x 1 = x 2 x 1 = x 2 to decide the function type.

⇒ x 1 2 = x 2 2 ⇒ x 1 2 = x 2 2

x 1 = ± x 2 x 1 = ± x 2

We see that " x 1 x 1 " is not exclusively equal to " x 2 x 2 ". Hence, given function is not one-one function, but many – one function. This conclusion is further emphasized by the intersection of a line parallel to x-axis, which intersects function plot at two points.

Figure 2: The line parallel to x-axis intersects function plot at two points.

Function type

Problem 3: A function f : R → R f : R → R is given by :

f x = x | x | f x = x | x |

Is the function one-one.

Solution :

Statement of the problem : Draw the plot of the function and see intersection of a line parallel to x-axis.

We observe from its plot that there is no line parallel to x-axis, which intersects the functions more than once. Hence, function is one-one.

Figure 3: The line parallel to x-axis intersects function plot at one point.

Function type

Problem 4: Determine whether greatest integer function is one-one function.

Solution : Statement of the problem : Draw the plot of the function and see intersection of a line parallel to x-axis.

Figure 4: The line parallel to x-axis intersects function plot at infinite pointes.

Function type

We have drawn one such line at y = 1. We see that this function value is valid for an interval of “x” given by 1≤x<2. Hence, greatest integer function is not one-one, but many –one function.

Problem 5: A function f : R → R f : R → R is given by :

f x = x 2 + 4 x + 30 x 2 − 8 x + 18 f x = x 2 + 4 x + 30 x 2 − 8 x + 18

Is the function one-one.

Solution :

Statement of the problem : The given function is a rational function. We have to determine function type.

We evaluate function for x =0. If f(x)=f(0) equation yields multiple values of “x”, then function in not one-one. Here,

⇒ f 0 = x 2 + 4 x + 30 x 2 − 8 x + 18 = 0 2 + 4 X 0 + 30 0 2 − 8 X 0 + 18 = 30 18 = 5 3 ⇒ f 0 = x 2 + 4 x + 30 x 2 − 8 x + 18 = 0 2 + 4 X 0 + 30 0 2 − 8 X 0 + 18 = 30 18 = 5 3

Now,

⇒ f x = f 0 ⇒ f x = f 0

⇒ f x = x 2 + 4 x + 30 x 2 − 8 x + 18 = 5 3 ⇒ f x = x 2 + 4 x + 30 x 2 − 8 x + 18 = 5 3

⇒ 3 x 2 + 12 x + 90 = 5 x 2 − 40 x + 90 ⇒ 2 x 2 − 52 x = 0 ⇒ 3 x 2 + 12 x + 90 = 5 x 2 − 40 x + 90 ⇒ 2 x 2 − 52 x = 0

⇒ x = 0, 26 ⇒ x = 0, 26

We see that f(0) = f(26). It means pre-images are not related to distinct images. Thus, we conclude that function is not one-one, but many-one.

Problem 6 : A continuous function f : R → R f : R → R is given by :

f x = x 2 + 4 x + 30 x 2 − 8 x + 18 f x = x 2 + 4 x + 30 x 2 − 8 x + 18

Determine increasing or decreasing nature of the function and check whether function is an injection?

Solution :

Statement of the problem : The rational function is a continuous function. Hence, we can determine its increasing or decreasing nature in its domain by examining derivative of the function.

⇒ f ′ x = x 2 − 8 x + 18 2 x + 4 − x 2 + 4 x + 30 2 x − 8 x 2 − 8 x + 18 2 ⇒ f ′ x = x 2 − 8 x + 18 2 x + 4 − x 2 + 4 x + 30 2 x − 8 x 2 − 8 x + 18 2

⇒ f ′ x = − 12 x 2 + 2 x − 26 x 2 − 8 x + 18 2 ⇒ f ′ x = − 12 x 2 + 2 x − 26 x 2 − 8 x + 18 2

The denominator is a square of a quadratic expression, which evaluates to a positive number. On the other hand, the discreminant of the quadratic equation in the numerator is :

⇒ D = 2 2 − 4 X 1 X − 26 = 4 + 104 = 108 ⇒ D = 2 2 − 4 X 1 X − 26 = 4 + 104 = 108

It means that derivative has different signs in the domain interval. Therefore, the function is a combination of increasing and decreasing nature in different intervals composing domain. Thus, function is not monotonic in the domain interval. Hence, we conclude that function is not an injection.

Problem 7: A function f : R → R f : R → R is given by :

f x = cos 3 x + 2 f x = cos 3 x + 2

Is the function one-one.

Solution :

Statement of the problem : We solve f x 1 = f x 2 f x 1 = f x 2 and see whether x 1 = x 2 x 1 = x 2 to decide the function type.

cos 3 x 1 + 2 = cos 3 x 2 + 2 cos 3 x 1 + 2 = cos 3 x 2 + 2

The basic solution is :

⇒ 3 x 1 + 2 = ± 3 x 2 + 2 ⇒ 3 x 1 + 2 = ± 3 x 2 + 2

General solution is obtained by adding integral multiples of the period of function, which is “2π” for cosine function :

⇒ 3 x 1 + 2 = 2 n π ± 3 x 2 + 2 ; n ∈ Z ⇒ 3 x 1 + 2 = 2 n π ± 3 x 2 + 2 ; n ∈ Z

For n = 1,

⇒ f 3 x 1 + 2 = f { 2 π ± 3 x 2 + 2 } ⇒ f 3 x 1 + 2 = f { 2 π ± 3 x 2 + 2 }

Thus, multiple pre-images are related to same image. Hence, given function is not one-one.

We can easily interpret from the plots of different trigonometric functions that they are not one-one functions. However, they are one-one in the subset of their domain. Such is the case with other functions as well. They are generally not one-one, but may reduce to one-one in certain interval(s).

Problem 8 : A function f : R → R f : R → R is given by :

f x = a x 2 + 6 x − 8 a + 6 x − 8 x 2 f x = a x 2 + 6 x − 8 a + 6 x − 8 x 2

Is the function one-one for a = 3?

Solution :

Statement of the problem : The given function is a rational function. Each of numerator and denominator functions is quadratic equation. In this case, solving f(a) = 0 for “x” reveals the nature of function for a =3.

⇒ f x = 3 x 2 + 6 x − 8 3 + 6 x − 8 x 2 = 0 ⇒ f x = 3 x 2 + 6 x − 8 3 + 6 x − 8 x 2 = 0

⇒ 3 x 2 + 6 x − 8 = 0 ⇒ 3 x 2 + 6 x − 8 = 0

⇒ x = - 6 ± 36 − 4 X 3 X - 8 6 = - 1 ± 33 3 ⇒ x = - 6 ± 36 − 4 X 3 X - 8 6 = - 1 ± 33 3

As “x” is not unique, the given function is not one-one function. We should emphasize here that solution of function when equated to zero is not a full proof method. In this particular case, it turns out that function value becomes zero for two values of “x”. In general, we should resort to techniques outlined in the beginning of the module to determine function type.