Using f is odd if f(-x) = -f(x) or even if f(-x) = f(x) for all real x, how do I
1)show that a polynomial P(x) that contains only odd powers of x is an odd function
2)show that if a polynomial P(x) contains both odd and even powders of x, then it is neither an odd nor an even function
Thanks in advance
You have the definition, so put -x in for x and see if it matches the odd or even definition. To be odd, all the terms have to change.
For instance f(x)= x + 3
That is not odd,nor even, because f(2)=5, and f(-2)=2, using your definitions
So to write the answer, I just use examples and prove it?
Thanks
To show that a polynomial P(x) that contains only odd powers of x is an odd function, we need to verify whether it satisfies the given definition. According to the definition, for a function to be odd, we need to check if f(-x) = -f(x) for all real values of x.
Let's consider a polynomial P(x) that contains only odd powers of x. We can write it in the form:
P(x) = a_1 * x + a_3 * x^3 + a_5 * x^5 + ...
Where a_1, a_3, a_5, ... are the coefficients of the polynomial.
Now, let's substitute -x for x in the polynomial:
P(-x) = a_1 * (-x) + a_3 * (-x)^3 + a_5 * (-x)^5 + ...
Since the powers of -x alternate in sign, we can simplify this expression as follows:
P(-x) = -a_1 * x - a_3 * x^3 - a_5 * x^5 - ...
Comparing P(-x) with -P(x), we can see that they are equivalent. Therefore, P(x) satisfies the condition f(-x) = -f(x), and hence, it is an odd function.
Now let's consider a polynomial P(x) that contains both odd and even powers of x. We can write it in the form:
P(x) = a_0 + a_1 * x + a_2 * x^2 + a_3 * x^3 + a_4 * x^4 + ...
Where a_0, a_1, a_2, a_3, a_4, ... are the coefficients of the polynomial.
We can repeat the same process and substitute -x for x in the polynomial:
P(-x) = a_0 + a_1 * (-x) + a_2 * (-x)^2 + a_3 * (-x)^3 + a_4 * (-x)^4 + ...
Simplifying this expression, we get:
P(-x) = a_0 - a_1 * x + a_2 * x^2 - a_3 * x^3 + a_4 * x^4 - ...
Comparing P(-x) with P(x), we can observe that they are not equivalent. Hence, P(x) does not satisfy either of the conditions f(-x) = -f(x) or f(-x) = f(x), and thus, it is neither an odd nor an even function.
In summary:
1) To show that a polynomial P(x) with only odd powers of x is an odd function, we substitute -x into P(x) and simplify, verifying if P(-x) = -P(x).
2) To show that a polynomial P(x) with both odd and even powers of x is neither an odd nor an even function, we substitute -x into P(x) and compare it with P(x), observing if they are equivalent or not.