Respuesta :
Answer:
(-inf,2]
Step-by-step explanation:
[tex](w \circ r)(x)=w(r(x))[/tex]
[tex]w(2-x^2)[/tex] I replaced r(x) with 2-x^2
[tex](2-x^2)-x[/tex] I replace the x in w(x)=x-2 with 2-x^2
[tex]-x^2-x+2[/tex]
You can graph this to find the range.
But since this is a quadratic (the graph is a parabola), I'm going to find the vertex to help me to determine the range.
The vertex is at x=-b/(2a). Once I find x, I can find the y that corresponds to it by using y=-x^2-x+2.
Comparing ax^2+bx+c to -x^2-x+2 tells me a=-1, b=-1, and c=2.
So the vertex is at x=1/(2*-1)=-1/2.
To find the y-coordinate that corresponds to that I will not plug in -1/2 in place of x into -x^2-x+2.
This gives me
-(-1/2)^2-(-1/2)+2
-1/4 + 1/2 +2
Find a common denominator which is 4.
-1/4 + 2/4 +8/4
8/4
2.
So the highest y value is 2 ( I know tha parabola is upside down because a=negative number)
That mean then range is 2 or less than 2.
So the answer an interval notation is (-inf,2]
