Question

What is the domain and range for `F(x) = -2(x + 3)² - 5`?

Answer 1

Domain: `D_f=R`
Range: `R_f=(-oo,-5]`

Explanation:

graph{-2(x+3)^2-5 [-11.62, 8.38, -13.48, -3.48]}

This is quadratic (polynomial) function so there aren't points of discontinuity and hence domain is `R` (set of real numbers).

`lim_(x->oo)(-2(x+3)^2-5)=-2(oo)^2-5=-2*oo-5=-oo-5=-oo`

`lim_(x->-oo)(-2(x+3)^2-5)=-2(-oo)^2-5=-2*oo-5=-oo-5=-oo`

However, function is bounded as you can see in graph so we have to find upper bound.

`F'(x)=-4(x+3)*1=-4(x+3)`
`F'(x_s)=0 <=> -4(x_s+3)=0 <=> x_s+3=0 <=> x_s=-3`

`AAx>x_s: F'(x)<0, F(x)` is decreasing

`AAx< x_s: F'(x)>0, F(x)` is increasing

So, `x_s` is maximum point and

`F_max=F(x_s)=F(-3)=-5`

Finally:

Domain: `D_f=R`
Range: `R_f=(-oo,-5]`