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

Question

1897 views around the world

You can reuse this answer
Creative Commons License

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]}

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