A piece wise function is used to illustrate functions of different input domain
The true statements are:
[tex](a)\ f(1) = 5[/tex]
[tex](d)\ f(2) = 4[/tex]
First, we test the options
[tex](a)\ f(1) = 5[/tex]
To calculate f(1), we make use of
[tex]f(x) = 5[/tex]
Because, the domain is [tex]x = 1[/tex]
So, we have:
[tex]f(x) = 5[/tex]
Substitute 1 for x
[tex]f(1) = 5[/tex]
Option (a) is true
[tex](b)\ f(5) = 1[/tex]
To calculate f(5), we make use of
[tex]f(x) =x^2[/tex]
Because, the domain is [tex]x > 1[/tex]
So, we have:
[tex]f(x) =x^2[/tex]
Substitute 5 for x
[tex]f(5) =5^2[/tex]
[tex]f(5) =25[/tex]
Option (b) is false
[tex](c)\ f(-2) = 4[/tex]
To calculate f(-2), we make use of
[tex]f(x) =2x[/tex]
Because, the domain is [tex]x < 1[/tex]
So, we have:
[tex]f(x) =2x[/tex]
Substitute -2 for x
[tex]f(-2) = 2 \times -2[/tex]
[tex]f(-2) = -4[/tex]
Option (c) is false
[tex](d)\ f(2) = 4[/tex]
To calculate f(2), we make use of
[tex]f(x) =x^2[/tex]
Because, the domain is [tex]x > 1[/tex]
So, we have:
[tex]f(x) =x^2[/tex]
Substitute 2 for x
[tex]f(2) = 2^2[/tex]
[tex]f(2) = 4[/tex]
Option (d) is true
Hence, the true statements are:
[tex](a)\ f(1) = 5[/tex]
[tex](d)\ f(2) = 4[/tex]
Read more about piece wise functions at:
https://brainly.com/question/2215108
