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

Aug 30, 2015

The domain of $F \left(x\right)$ is $\left(- \infty , \infty\right)$.

The range of $F \left(x\right)$ is $\left(- \infty , 6 \sqrt[3]{4} - 1\right) \approx \left(- \infty , 8.5244\right)$

#### Explanation:

$F \left(x\right)$ is well defined for all $x \in \mathbb{R}$, so the domain is $\mathbb{R}$ or $\left(- \infty , + \infty\right)$ in interval notation.

$F ' \left(x\right) = - 2 {x}^{3} + 8 = - 2 \left({x}^{3} - 4\right)$

So $F ' \left(x\right) = 0$ when $x = \sqrt[3]{4}$. This is the only Real zero of $F ' \left(x\right)$, so the only turning point of $F \left(x\right)$.

$F \left(\sqrt[3]{4}\right) = - \frac{1}{2} {\left(\sqrt[3]{4}\right)}^{4} + 8 \sqrt[3]{4} - 1$

$= - 2 \sqrt[3]{4} + 8 \sqrt[3]{4} - 1 = 6 \sqrt[3]{4} - 1$

Since the coefficient of ${x}^{4}$ in $F \left(x\right)$ is negative, this is the maximum value of $F \left(x\right)$.

So the range of $F \left(x\right)$ is $\left(- \infty , 6 \sqrt[3]{4} - 1\right) \approx \left(- \infty , 8.5244\right)$

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