Question

How do I find the upper bound of a polynomial?

Answer 1

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(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_1 x + a_0`

you need to find the zeros of the derivative:

`f'(x) = n a_n x^(n-1) + (n-1) a_(n-1) x^(n-2) + ... + a_1`

and evaluate `f(x)` 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(x) = -x^4+4x^3-2x^2-4x+1`

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

We find:

`f'(x) = -4x^3+12x^2-4x-4`

`= -4(x^3-3x^2+x+1)`

`= -4(x-1)(x^2-2x-1)`

`= -4(x-1)(x-1-sqrt(2))(x-1+sqrt(2))`

So evaluate `f(x)` for each of the three zeros of `f'(x)`.

`f(x) = -x^4+4x^3-2x^2-4x+1=1-x(x((x-4)x+2)+4)`

So:

`f(1) = -1+4-2-4+1 = -2`

`f(1+sqrt(2)) =1-(1+sqrt(2))((1+sqrt(2))(((1+sqrt(2))-4)(1+sqrt(2))+2)+4) = 2`

`f(1-sqrt(2)) =1-(1-sqrt(2))((1-sqrt(2))(((1-sqrt(2))-4)(1-sqrt(2))+2)+4) = 2`

So the upper bound of `f(x)` is `2`

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