Question

The function f is defined by f(x) = {(x^2,0≤x≤3)(3x,3≤x≤10)

\(f(x) = \begin{cases} x^2, 0≤x≤3\\ 3x,3≤x≤10 \end{cases}\)

The relation g is defined by

 \(g(x) = \begin{cases} x^2, 0≤x≤2\\ 3x,2≤x≤10 \end{cases}\) 

Show that f is a function and g is not a function.

Answer 1

Given,

f(x) = {(x^2,0≤x≤3)(3x,3≤x≤10) and

\(g(x) = \begin{cases} x^2, 0≤x≤2\\ 3x,2≤x≤10 \end{cases}\)

Let us first show that f is a function. 

When 0 ≤ x ≤ 3, 

f(x) = x². 

The function x² associates all the numbers 0 ≤ x ≤ 3 to unique numbers in R. 

Hence, 

The images of {x ∈ Z: 0 ≤ x ≤ 3} exist and are unique. 

When 3 ≤ x ≤ 10, 

f(x) = 3x. 

The function x² associates all the numbers 3 ≤ x ≤ 10 to unique numbers in R. 

Hence, 

The images of {x ∈ Z: 3 ≤ x ≤ 10} exist and are unique. 

When x = 3, 

Using the first definition, we have 

f(3) = 3² = 9 

When x = 3, 

Using the second definition, we have 

f(3) = 3(3) = 9 

Hence, 

The image of x = 3 is also distinct. 

Thus, 

As every element of the domain has an image and no element has more than one image, f is a function. 

Now, 

Let us show that g is not a function. 

When 0 ≤ x ≤ 2, g(x) = x². 

The function x² associates all the numbers 0 ≤ x ≤ 2 to unique numbers in R. 

Hence, 

The images of {x ∈ Z: 0 ≤ x ≤ 2} exist and are unique. 

When 2 ≤ x ≤ 10, g(x) = 3x. 

The function x² associates all the numbers 2 ≤ x ≤ 10 to unique numbers in R. 

Hence, 

The images of {x ∈ Z: 2 ≤ x ≤ 10} exist and are unique. 

When x = 2, 

Using the first definition, we have 

g(2) = 2² = 4 

When x = 2, 

Using the second definition, we have 

g(2) = 3(2) = 6 

Here, 

The element 2 of the domain is associated with two elements distinct elements 4 and 6. 

Thus, 

g is not a function.