The graph of the function f(x) = –(x + 3)(x – 1) is shown below. what is true about the domain and range of the function? the domain is all real numbers less than or equal to 4, and the range is all real numbers such that –3 ≤ x ≤ 1. the domain is all real numbers such that –3 ≤ x ≤ 1, and the range is all real numbers less than or equal to 4. the domain is all real numbers, and the range is all real numbers less than or equal to 4. the domain is all real numbers less than or equal to 4, and the range is all real numbers.

The domain is all real numbers, and the range is all real numbers less than or equal to 4.

Step-by-step explanation:

The domain is the horizontal extent. The function is a polynomial. The domain of any polynomial is "all real numbers".

The range is the vertical extent. The function is an even-degree polynomial with a negative x² coefficient, so it will open downward. That means it has an absolute maximum that restricts its range. Here, that maximum is 4, so the range is ...

... all real numbers less than or equal to 4.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!