Hi I am new to the forum and I am looking for someone to confirm my work.
Show if the function is even, odd, or neither
f(x) = x^3 + (x)(sin x)^2
First I solve for f(-x)
f(-x) = (-x)^3 + (-x)[sin(-x)]^2
f(-x) = (-x)^3 + (-x)(sin -x)(sin -x) sin of negative x is the same as the negative of sin of x
f(-x) = (-x)^3 + (-x)(- sin x)(- sin x) negative sin of x quantity squared, the negatives cancel
f(-x) = (-x)^3 + (-x)(sin x)^2 factor out the negative
f(-x) = -[(x^3) + (x)(sin x)^2] this shows that f(-x) = - f(x) therefore the function is odd
I am very certain this is right. I would appreciate if anyone could confirm.
Thanks!
Show if the function is even, odd, or neither
f(x) = x^3 + (x)(sin x)^2
First I solve for f(-x)
f(-x) = (-x)^3 + (-x)[sin(-x)]^2
f(-x) = (-x)^3 + (-x)(sin -x)(sin -x) sin of negative x is the same as the negative of sin of x
f(-x) = (-x)^3 + (-x)(- sin x)(- sin x) negative sin of x quantity squared, the negatives cancel
f(-x) = (-x)^3 + (-x)(sin x)^2 factor out the negative
f(-x) = -[(x^3) + (x)(sin x)^2] this shows that f(-x) = - f(x) therefore the function is odd
I am very certain this is right. I would appreciate if anyone could confirm.
Thanks!