What is the domain and range of #F (x)= -1/2 x^4+8x-1#?

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Answer 1 · 2015-08-30T07:16:07

The domain of #F(x)# is #(-oo, oo)#.

The range of #F(x)# is #(-oo, 6root(3)(4)-1) ~~ (-oo, 8.5244)#

#F(x)# is well defined for all #x in RR#, so the domain is #RR# or #(-oo, +oo)# in interval notation.

#F'(x) = -2x^3+8 = -2(x^3-4)#

So #F'(x) = 0# when #x = root(3)(4)#. This is the only Real zero of #F'(x)#, so the only turning point of #F(x)#.

#F(root(3)(4)) = -1/2(root(3)(4))^4+8root(3)(4)-1#

#=-2root(3)(4)+8root(3)(4)-1 = 6root(3)(4)-1#

Since the coefficient of #x^4# in #F(x)# is negative, this is the maximum value of #F(x)#.

So the range of #F(x)# is #(-oo, 6root(3)(4)-1) ~~ (-oo, 8.5244)#

graph{ -1/2x^4+8x-1 [-9.46, 10.54, -1, 9]}