# How do you find the domain and range of 5x^4+x^3-19x^2-9x+8?

##### 1 Answer
May 8, 2018

The domain is $\mathbb{R}$. Here's a rough sketch of how to find the range...

#### Explanation:

Given:

$f \left(x\right) = 5 {x}^{4} + {x}^{3} - 19 {x}^{2} - 9 x + 8$

Note that this is a polynomial, so has domain the whole of $\mathbb{R}$.

It is of even degree ($4$) with positive leading coefficient, so has range of the form $\left[k , \infty\right)$ for some $k$, since it is continuous with at least one global minimum and no upper limit.

To find the minima, we can see where the derivative is zero.

$f ' \left(x\right) = 20 {x}^{3} + 3 {x}^{2} - 38 x - 9$

This cubic has $3$ real irrational zeros, which will correspond to two minima and one local maximum of $f \left(x\right)$.

Here are $f \left(x\right)$ and $f ' \left(x\right)$ plotted together. You can see that the cubic $f ' \left(x\right)$ has a zero at roughly $x = 1.4$ corresponding to the minimum of $f \left(x\right)$

graph{(y-(5x^4+x^3-19x^2-9x+8))(y-(20x^3+3x^2-38x-9)) = 0 [-2.5, 2.5, -35, 25]}

$f ' \left(x\right) = 0$ can be solved algebraically using a trigonometric substitution.

Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 20$, $b = 3$, $c = - 38$ and $d = - 9$, so we find:

$\Delta = 12996 + 4389760 + 972 - 874800 + 369360 = 3898288$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 400 f ' \left(x\right) = 8000 {x}^{3} + 1200 {x}^{2} - 15200 x - 3600$

$= {\left(20 x + 1\right)}^{3} - 763 \left(20 x + 1\right) - 2838$

$= {t}^{3} - 763 t - 2838$

where $t = \left(20 x + 1\right)$

Trigonometric substitution

To solve ${t}^{3} - 763 t - 2638 = 0$ the idea is to substitute $t = k \cos \theta$ with $k$ chosen so that the resulting cubic contains $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$

Let $k = \sqrt{\frac{4 \cdot 763}{3}} = \frac{2}{3} \sqrt{2289}$

Then:

$0 = {t}^{3} - 763 t - 2638$

$\textcolor{w h i t e}{0} = k \left({k}^{2} {\cos}^{3} \theta - 763 \cos \theta\right) - 2638$

$\textcolor{w h i t e}{0} = \frac{763 k}{3} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 2638$

$\textcolor{w h i t e}{0} = \frac{763 k}{3} \cos 3 \theta - 2638$

So:

$\cos 3 \theta = \frac{2638 \cdot 3}{763 k}$

$\textcolor{w h i t e}{\cos 3 \theta} = \frac{2638 \cdot 3}{763 \cdot \frac{2}{3} \sqrt{2289}}$

$\textcolor{w h i t e}{\cos 3 \theta} = \frac{11871}{763 \cdot \sqrt{2289}}$

$\textcolor{w h i t e}{\cos 3 \theta} = \frac{3957}{582169} \sqrt{2289}$

Hence:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{3957}{582169} \sqrt{2289}\right) + 2 n \pi$

This yields distinct values:

${t}_{n} = \frac{2}{3} \sqrt{2289} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3957}{582169} \sqrt{2289}\right) + \frac{2 n \pi}{3}\right)$

for $n = 0 , 1 , 2$

and hence:

${x}_{n} = \frac{1}{20} \left(- 1 + \frac{2}{3} \sqrt{2289} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3957}{582169} \sqrt{2289}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$

In particular:

${x}_{0} \approx 1.41057$

So the range of the given function is:

$\left[f \left({x}_{0}\right) , \infty\right)$

where:

${x}_{0} = \frac{1}{20} \left(- 1 + \frac{2}{3} \sqrt{2289} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3957}{582169} \sqrt{2289}\right)\right)\right)$