Question

Can you please explain to me which functions are odd?

f(x)=-1/2x^4+5

f(x)=-8x^3+5x

f(x)=-4/x^3-x+1

f(x)=x^5/x^4-1

f(x)=-sqrt2x

f(x)=3sqrtx -x^3

thank you!

Answer 1

A function is odd if f(-x)=-f(x).
So given a function, to check whether it is odd or not, we calculate f(-x). If it is equal to -f(x), then the function is odd; if not, it is not odd.



`\displaystyle{f(x)= -(1)/(2)x^4+5`


`\displaystyle{f(-x)= -(1)/(2)(-x)^4+5= -(1)/(2)x^4+5\\eq-f(x)`

thus the function is not odd



`\displaystyle{f(x)=-8x^3+5x`


`\displaystyle{f(-x)=-8(-x)^3+5(-x)=8x^3-5x=-(-8x^3+5x)=-f(x)`

thus the function is odd




`\displaystyle{f(x)=- (4)/(x^3)-x+1`


`\displaystyle{f(-x)=- (4)/((-x)^3)-(-x)+1= (4)/(x^3)+x+1\\eq-f(x)`

thus the function is not odd




`\displaystyle{f(x)= (x^5)/(x^4-1)`


`\displaystyle{f(-x)= ((-x)^5)/((-x)^4-1)= (-x^5)/(x^4-1)=- (x^5)/(x^4-1)=-f(x)`

thus the function is odd




`\displaystyle{f(x)=-√(2x)`


`\displaystyle{f(-x)=-√(2(-x))= -√(-2x)`

In this particular case f(x) and f(-x) can both exist only for x=0 (because one of them is certainly negative). Thus the function is not odd



`\displaystyle{f(x)=3 √(x) -x^3`

similarly to the previous case, the Domain of f is [0, infinity) and f(-x) cannot be calculated except for x=0. So the function is not odd.