Find the bounds on the real zeros of the following function. Please show all of your work.
f(x)= x^4 + 5x^3 + 6x^2 - 4.
my question is, what does "bound" mean?
is this the same as the vertical asymptote, if not how is different?
The roots of your function are not rational
I used this source to approximate
http://xrjunque.nom.es/precis/rootfinder.aspx
(enter the expression x^4 + 5x^3 + 6x^2 - 4 in the input box )
x = appr. - 3.29 and x = appr. 0.64
with two additional complex roots.
I don't know what "bounds of the real zeros" means, have never seen a question worded like that before.
To find the bounds on the real zeros of the function f(x) = x^4 + 5x^3 + 6x^2 - 4, we can use Descartes' Rule of Signs.
1. Count the number of sign changes in the coefficients of the polynomial. We have:
Number of sign changes in f(x) = x^4 + 5x^3 + 6x^2 - 4:
- From +x^4 to +5x^3: 1 sign change
- From +5x^3 to +6x^2: 0 sign changes
- From +6x^2 to -4: 1 sign change
So, there are 2 sign changes in the coefficients.
2. Determine the possible number of positive real zeros.
The number of positive real zeros can be equal to the number of sign changes in f(x) or less than that by an even number. In this case, the number of sign changes is 2, so there are either 2 positive real zeros or 0 positive real zeros.
3. Determine the possible number of negative real zeros.
The number of negative real zeros can be determined by finding f(-x) and counting the number of sign changes in its coefficients.
Let g(x) = f(-x) = (-x)^4 + 5(-x)^3 + 6(-x)^2 - 4 = x^4 - 5x^3 + 6x^2 - 4.
Number of sign changes in g(x) = x^4 - 5x^3 + 6x^2 - 4:
- From +x^4 to -5x^3: 1 sign change
- From -5x^3 to +6x^2: 2 sign changes
- From +6x^2 to -4: 1 sign change
So, there are 4 sign changes in the coefficients.
The number of negative real zeros can be either 4 or a number less than that by an even number. In this case, the number of sign changes is 4, so there are either 4 negative real zeros, 2 negative real zeros, or 0 negative real zeros.
Therefore, the bounds on the real zeros of f(x) = x^4 + 5x^3 + 6x^2 - 4 are:
- There are at most 2 positive real zeros.
- There are at most 4 negative real zeros.