Respuesta :
Answer:
Domain: (-∞,∞) or -∞ <[tex]x[/tex]<∞
Range: [1,∞) or [tex]y\geq 1[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=x^2-2x+2[/tex]
To find range and domain of the function.
Solution:
The given function is a second degree function as the highest exponent of the variable is 2. Thus, it is a quadratic function.
For all quadratic functions the domain is a set of all real numbers. So, the domain can be given as: (-∞,∞) or -∞ <[tex]x[/tex]<∞
In order to find the range of the quadratic function, we will first determine if the function has a minimum point or the maximum point.
For a quadratic equation : [tex]ax^2+bx+c[/tex]
1) If [tex]a>0[/tex] the function will have a minimum point.
2) If [tex]a<0[/tex] the function will have a maximum point.
For the given function: [tex]f(x)=x^2-2x+2[/tex]
[tex]a=1[/tex] which is greater than 0, and hence it will have a minimum point.
To find the range we will find the coordinates of the minimum point or the vertex of the function.
The x-coordinate [tex]h[/tex] of the vertex of a quadratic function is given by :
[tex]h=\frac{-b}{2a}[/tex]
Thus, for the function:
[tex]h=\frac{-(-2)}{2(1)}[/tex]
[tex]h=\frac{2}{2}[/tex] [Two negatives multiply to get a positive]
[tex]h=1[/tex]
To find y-coordinate of the vertex can be found out by evaluating [tex]f(h)[/tex].
[tex]k=f(h)[/tex]
Thus, for the function:
[tex]k=f(1)=(1)^2-2(1)+2[/tex]
[tex]k=f(1)=1-2+2[/tex]
[tex]k=f(1)=1[/tex]
Thus, the minimum point of the function is at (1,1).
thus, range of the function is:
[1,∞) or [tex]y\geq 1[/tex]
