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You need to demonstrate that the function is not injective, hence, you need to prove that at least one element in the range of function is the image of more than one elements from domain of function.
You need to start with the assumption that `f(x_1) = f(x_2)` , such that:
`f(x_1) = f(x_2) => 2x_1^5+3x_1^3+7 = 2x_2^5+3x_2^3+7`
Reducing like terms yields:
`2x_1^5+3x_1^3 = 2x_2^5+3x_2^3`
`2x_1^5 - 2x_2^5 = 3x_2^3 - 3x_1^3`
`2(x_1^5 - x_2^5) = -3(x_1^3 - x_2^3)`
You may use the following formula, such that:
`x_1^n - x_2^n = (x_1 - x_2)(x_1^(n-1) + x_1^(n-2)*x_2 + ... + x_2^(n-1))`
Reasoning by analogy, yields:
`2(x_1 - x_2)(x_1^4 + x_1^3*x_2 + x_1^2*x_2^2 + x_1*x_2^3 + x_2^4) = -3(x_1 - x_2)(x_1^2 + x_1*x_2 + x_2^2)`
Reducing duplicate factors, yields:
`2(x_1^4 + x_1^3*x_2 + x_1^2*x_2^2 + x_1*x_2^3 + x_2^4) = -3(x_1^2 + x_1*x_2 + x_2^2) => x_1 != x_2`
Hence, you have started with the assumption that the images `f(x_1)` and `f(x_2)` are equal and you have reached to the conclusion that `x_1 != x_2` , hence, since because the image `f(x_1)=f(x_2)=y` corresponds to two different elements in domain, `x_1` and `x_2` , such that`f(x_1) = f(x_2) => x_1 != x_2` , yields that the function is not injective.
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