# Determining if polynomial function

1. Sep 17, 2011

### Nelo

1. The problem statement, all variables and given/known data

We know that a polynomial function is anything with a positive exponent and a rational number.

a) - / x^3 -4

d) 3x^-1 - 11

g) y= [sqrt of term]3x^2 -5x

2. Relevant equations

3. The attempt at a solution

c) 1 / x^3 - 4
(do we use exponents to verify ?)
i
e) 1^1 / x^3 -4^1
1-3 = -2 , 1-1 = 0
Therefore there is a negetive exponent on the x^3, is that why?

4d) y= 3^x-1 - 11
(how do u know this isnt?)

g) [sqrtofentireterm] 3x^3-5x

Why is that not a polynomial function?

Thnx 4 the help!

2. Sep 17, 2011

### HallsofIvy

Staff Emeritus
No, we don't know that! That is wrong. A polynomial function can has the variable, x, only to positive integer exponents. The coefficients can be any numbers, not just rational numbers.

what does the "-/" mean? Is that an attempt at a squareroot symbol? Better would be "sqrt(x^3- 4) which is the same as (x^3- 4)^(1/2). That's not a polynomial because it has a fractional exponent.

negative exponent

again, sqrt= exponent 1/2.

is that 1/(x^3- 4) or (1/x^3)- 4?
In either case, there is a negative exponent, (x^3- 4)^(-1) or x^(-3)- 4.

I hav no clue what you are doing here.

3^(x- 1) does not have x to a power.

Again, because of the square root= 1/2 power.

3. Sep 17, 2011

### Nelo

Therefore there is a negetive exponent on the x^3, is that why?

4d) y= 3^x-1 - 11
(how do u know this isnt?)

3^(x- 1) does not have x to a power

Why do you mean does not have x to a power??

c) 1 / x^3 - 4
(do we use exponents to verify ?)

is that 1/(x^3- 4) or (1/x^3)- 4?
In either case, there is a negative exponent, (x^3- 4)^(-1) or x^(-3)- 4.

How is there a negetive exponent there... how do u verify? There is no braackets anywhere on my page so i wrote it as written. Do u evalue the exponents in the divsion and then it comes out as a negetive exponent?

4. Sep 18, 2011

### eumyang

Do you know the properties of exponents? Specifically, this one:
$$a^{-n} = \frac{1}{a^n}, a \ne 0$$
If not, you'll need to review them.