Verify if the function is odd or even? y=17x^3-12x^2
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
A function is odd if f(-x) = -f(x) and even if f(-x) = f(x)
Here we have f(x) = y= 17x^3 - 12x^2
f(-x) = -17x^3 - 12x^2
f(x) = 17x^3 - 12x^2
-f(x) = -17x^3 + 12x^2
So we see that f(-x) = -17x^3 - 12x^2 is neither equal to f(x) nor equal to -f(x).
Therefore the function is neither even nor odd.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
A function is even if f(-x) = f(x).
In other words, plugging in a number will be the same as plugging in the negative value of the same number. The function is not changing.
We'll analyze the given function, replacing each x by -x.
f(-x) = 17(-x)^3 - 12(-x)^2
We'll compute raising -x to the 3rd and 2nd powers and we'll get:
(-x)^3 = (-x)(-x)(-x) = x^2*(-x) = -x^3
f(-x) = -17x^3 - 12x^2
So we can see that:
f(-x) is not equal to f(x) which means that the function f(x) is not an even function.
We'll check if the function is odd;
f(-x) = -f(x)
f(-x) = -17x^3 - 12x^2
If we'll factorize by -1 we'll get:
f(-x) = -(17x^3 + 12x^2)
The expression inside brackets is not the function f(x).
The given function is nor odd neither even function.