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

1 Answer
May 8, 2018

Answer:

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))))#