Question

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

Answer 1

The domain is `RR`. Here's a rough sketch of how to find the range...

Explanation:

Given:

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

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

It is of even degree (`4`) with positive leading coefficient, so has range of the form `[k, oo)` 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'(x) = 20x^3+3x^2-38x-9`

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

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

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'(x) = 0` can be solved algebraically using a trigonometric substitution.

Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

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

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

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

`=t^3-763t-2838`

where `t=(20x+1)`

Trigonometric substitution

To solve `t^3-763t-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((4 * 763)/3) = 2/3 sqrt(2289)`

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 `n = 0, 1, 2`

and hence:

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

for `n = 0, 1, 2`

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))))`