# How do I find the upper bound of a polynomial?

May 17, 2016

See explanation...

#### Explanation:

If the term of highest degree is of odd degree or has a positive coefficient, then there is no upper bound (unless you are being asked for the upper bound over an interval).

Otherwise, in the general case of:

$f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

you need to find the zeros of the derivative:

$f ' \left(x\right) = n {a}_{n} {x}^{n - 1} + \left(n - 1\right) {a}_{n - 1} {x}^{n - 2} + \ldots + {a}_{1}$

and evaluate $f \left(x\right)$ at those zeros.

The highest of these values will be the upper bound of the polynomial.

Example

What is the upper bound of the following polynomial?

$f \left(x\right) = - {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 4 x + 1$

Note that this polynomial has even degree and a negative leading coefficient, so does have an upper bound.

We find:

$f ' \left(x\right) = - 4 {x}^{3} + 12 {x}^{2} - 4 x - 4$

$= - 4 \left({x}^{3} - 3 {x}^{2} + x + 1\right)$

$= - 4 \left(x - 1\right) \left({x}^{2} - 2 x - 1\right)$

$= - 4 \left(x - 1\right) \left(x - 1 - \sqrt{2}\right) \left(x - 1 + \sqrt{2}\right)$

So evaluate $f \left(x\right)$ for each of the three zeros of $f ' \left(x\right)$.

$f \left(x\right) = - {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 4 x + 1 = 1 - x \left(x \left(\left(x - 4\right) x + 2\right) + 4\right)$

So:

$f \left(1\right) = - 1 + 4 - 2 - 4 + 1 = - 2$

$f \left(1 + \sqrt{2}\right) = 1 - \left(1 + \sqrt{2}\right) \left(\left(1 + \sqrt{2}\right) \left(\left(\left(1 + \sqrt{2}\right) - 4\right) \left(1 + \sqrt{2}\right) + 2\right) + 4\right) = 2$

$f \left(1 - \sqrt{2}\right) = 1 - \left(1 - \sqrt{2}\right) \left(\left(1 - \sqrt{2}\right) \left(\left(\left(1 - \sqrt{2}\right) - 4\right) \left(1 - \sqrt{2}\right) + 2\right) + 4\right) = 2$

So the upper bound of $f \left(x\right)$ is $2$

graph{-x^4+4x^3-2x^2-4x+1 [-3.79, 6.21, -2.48, 2.52]}