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

##### 1 Answer

#### Answer:

The domain is

#### Explanation:

Given:

#f(x) = 5x^4+x^3-19x^2-9x+8#

Note that this is a polynomial, so has domain the whole of

It is of even degree (

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

#f'(x) = 20x^3+3x^2-38x-9#

This cubic has

Here are

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

**Discriminant**

The discriminant

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example,

#Delta = 12996+4389760+972-874800+369360 = 3898288#

Since

**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=400f'(x)=8000x^3+1200x^2-15200x-3600#

#=(20x+1)^3-763(20x+1)-2838#

#=t^3-763t-2838#

where

**Trigonometric substitution**

To solve

Let

Then:

#0 = t^3-763t-2638#

#color(white)(0) = k(k^2 cos^3 theta - 763 cos theta)-2638#

#color(white)(0) = (763k)/3 (4 cos^3 theta - 3 cos theta)-2638#

#color(white)(0) = (763k)/3 cos 3 theta-2638#

So:

#cos 3 theta = (2638 * 3)/(763 k)#

#color(white)(cos 3 theta) = (2638 * 3)/(763 * 2/3 sqrt(2289))#

#color(white)(cos 3 theta) = 11871/(763 * sqrt(2289))#

#color(white)(cos 3 theta) = 3957/582169 sqrt(2289)#

Hence:

#3 theta = +-cos^(-1)(3957/582169 sqrt(2289))+2npi#

This yields distinct values:

#t_n = 2/3 sqrt(2289) cos(1/3cos^(-1)(3957/582169 sqrt(2289))+(2npi)/3)#

for

and hence:

#x_n = 1/20(-1+2/3 sqrt(2289) cos(1/3cos^(-1)(3957/582169 sqrt(2289))+(2npi)/3))#

for

In particular:

#x_0 ~~ 1.41057#

So the range of the given function is:

#[f(x_0), oo)#

where:

#x_0 = 1/20(-1+2/3 sqrt(2289) cos(1/3cos^(-1)(3957/582169 sqrt(2289))))#