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

1 Answer
Sep 16, 2015

Answer:

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]#