Why would f(x) =pi/4x^3-sqrt2x+4 be classified as a polynomial?


How would you determine the domain and range of y=x^4-4x^2?

Assuming that the zeros of the function y=x^3+mx^2+nx+2 are integers how would you determine the values of m and n??

Thank you.

How would you determine the domain and range of y=x^4-4x^2?

y = (x^4) - 4*(x^2)
y = (x^2) * (x^2 - 4); this says there zeroes at 2 and -2, and two zeroes at x=0.

y' = 4*(x^3) - 8x; set equal to zero, and there are zeroes at 0, +SQRT(2), -SQRT(2).
These are potential minimum and maximum points.

y'' = 8*(x^2) - 8 = 8*(x^2 - 1).
y''(0) < 0, so 0 is a relative maximum; it is not an absolute maximum.
y''(+SQRT2) > 0, so +SQRT2 is a relative minimum.
y''(-SQRT2) > 0, so -SQRT2 is a relative minimum.

domain: x can be any real number.

range: As x gets infintely large, y gets infinitely large. But y is a
minimum at x = + or -SQRT2:
y = (x^4) - 4*(x^2) = (4) - 4*(2) = -4. So -4<y< infinity.

Why would f(x) =pi/4x^3-sqrt2x+4 be classified as a polynomial?
Polynomial is any function of x where x is only taking forms of its non-negative whole number powers, and coefficients that stand next to such "forms" of x are any given number.

Examples of some polynomials:
f(x) = x^2 - x + 1
f(x) = (29/3)*x^3
f(x) = (pi/4 - sqrt(2*pi-18i))*x^192 - (tg(pi/3)-e^45)*x^2 + (2/35)

Examples of some non-polynomials:
g(x) = sqrt(x) <- here the power of x is (1/2) which is not a whole number
g(x) = e^x <- x is in the power, and not the base
g(x) = x^2 + x - (1/x) <- the last part is -(x^(-1)), and -1 is not a non-negative number

In your example you have:
f(x) = (pi/4)*x^3 - sqrt(2)*x + 4
This is a polynomial since x comes in the forms of powers to the 0, 1 and 3, which are all non-negative whole numbers.

How would you determine the domain and range of y=x^4-4x^2?
I want to comment a bit on how to calculate the domain of a function. Domain of a function is a set of numbers for which you can calculate the value of a function.

For example, if you have f(x) = x, it's domain is the whole set of real numbers, because for every real number x, you can calculate how much is f(x).
For example:
f(5) = 5
f(246928641982612) = 246928641982612
f(-97097353856398146248) = -97097353856398146248

Let's take a look at f(x) = x^2 - 2
Again, it's domain is the whole set of real numbers.
Let's take a look:
f(2) = 2
f(-8) = 62
f(1598636) = 2555637060494
f(-98690376094376) = 9739790333649381859658829374

For another example, let's look at f(x) = x / (x - 2)
Now the domain of that function is not the whole set of real numbers, because you cannot calculate f(2), since f(2) = 2 / (2 - 2) = 2 / 0, and we do not know how to divide by zero.
So domain of that function is the whole set of real numbers, without number 2.

Another example is f(x) = (x - 3) / (x^2 - 1)
Domain is the whole set of real numbers, without numbers 1 and -1. Try to find out for yourself why it's so.

That's awesome, thank you so much.

Your polynomial x^3 + mx^2 + nx + 2 can be written as the product of three terms (like all 3rd order polynomials).

(x+a)(x+b)(x+c)

Expand this, and you get x^3 + (ab+ac+bc)x^2 + (a+b+c)x + abc

So, abc=2, a+b+c=n, ab+ac+bc=m. Any more information? Yes. N and m are both integers. It follows (I think) that a, b and c all have to be integers as well. So, how many combinations of three integers can multiply out to 2 (order is not important):

1 x 1 x 2
-1 x -1 x 2
1 x -1 x -2

Now plug these in and you get the following pairs for (m,n): (5,4), (-3,0), (-1,-2).

For x^3 + mx^2 + nx + 2 , how would you solve it out if you knew the zeros were -1, 2 and 1? Like how can you substitute those numbers in to solve for m and n?

Thank you.

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

like that.