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Given $X = \{-1, 0, 1\}$ and $f(x) = 2x^4$. We calculate the image of each element:
For $x = -1$: $f(-1) = 2(-1)^4 = 2(1) = 2$
For $x = 0$: $f(0) = 2(0)^4 = 2(0) = 0$
For $x = 1$: $f(1) = 2(1)^4 = 2(1) = 2$
A function is one-one if distinct elements have distinct images. Here, $f(-1) = 2$ and $f(1) = 2$. Since two distinct elements in the domain map to the same element in the codomain, the function is not one-one.
A function is onto if the range equals the codomain. The range of $f$ is $\{0, 2\}$. The codomain $Y$ is given as $\{0, 2\}$. Since Range = Codomain, every element in $Y$ has a pre-image in $X$. Therefore, the function is onto.
Final Answer: The function is not one-one but it is onto.
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