Finding the domain of a square root function

It has been a long time since I've posted. I apologize for this inconvenience. I will also attempt to answer questions. I enrolled at LCC. It has been 5 years since I've taken a Calculus course, so I've gone back to college Algebra. Since I've gone back to College Algebra, I've reviewed and learned some new things, here is an interesting problem: Solve for x.

f(x)=sqrt x^4-26x^2+25

The first thing you would do is: FACTOR the polynomial into linear factors. From the leading-term test we know that f(x) has at most 4 asymptotes since it is of the fourth degree. Also since it is of the fourth degree we know the beginning and end behaviour of the graph. We know it is concave up since the leading term is positive and that it is a parabola shape since the degree is even. But now we have to factor the polynomial into linear factors. The factor theorem states that:

For a Polynomial of f(x) if f(c)=0 , then x-c is a factor of f(x). Simply put if f(3) =0 then (x-3) is a factor for an example. We can factor this by stating the possible rational zeros of the function. The possible rational zeros of the function are P/Q. Which is constant term over coefficient of leading term. Therefore the possible rational zeros are: +1,-1,+5,-5,+25,-25. -1 happens to be a zero, therefore we find the 3rd degree equivalent polynomial with synthetic division

-1

1 0 -26 0 25

-1 1 25 -25

1 -1 -25 25 0

Therefore these are our coefficients to our 3rd degree polynomial. x^3-x^2-25x+25=0

-5 is a zero of this equivalent polynomial. Therefore use synthetic division again to get a second degree polynomial.

-5

1 -1 -25 25

-5 30 -25

1 -6 5 0


Our second degree polynomial is of the sort. x^2-6x+5=0

This of couse can easily be factored! into (x-1)(x-5)

Therefore our zeros of the polynomial function are -1,-5, 1 and 5.

So our linear factors are (x-1)(x+1)(x-5)(x+5)= x^4-26x^2+25.

I know what some of you are thinking.....So what does this have to do with square root of f(x). It has a lot to do with it! First we need to factor the polynomial and since that all square roots are postive we need to set the linear factors greater than or equal to zero.

So (x-1)(x+1)(x-5)(x+5)=0

Set it equal to zero, we find the critical points 1,-1,5,-5

therefore we have the intervals of:

(-inf, -5) U (-5,-1) U (-1,1) U (1,5) U (5,inf).

+ - + - +


We find where the graph is greater or equal to zero. It is with 3 intervals!

So our answer to finding the square root of f(x) is the intervals:

(-inf, -5) U (-1,1) U (5,inf)

Punch this function into a calculator---- very interesting piecewise function.

How about a function like this:

x^4+6x^3+11x^2+6x+1. The square root function is not much different than the original and it is hard to find the zeros to this function. Do you know why? By Desecrates rule of signs there are no sign variations therefore there are zero positive rational zeros. To find the negative rational zeros you try f(-x) and f(-x)= (-x)^4+6(-x)^3+11(-x)^2+6(-x)+1. Hence f(-x) is x^4-6x^3+11x^2-6x+1. There are 4 variations of this. So there are either 4 or 4-2=1 or 4-4= 0 variations. I plugged this equation into my voyage 200 calculator and the zeros are irrational zeros. So if anyone knows how to find the zeros to this function let me know. There are only 2 and they are both irrational.