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Prove that the function

Prove that the function-example-1
User Mike Fikes
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Answer:

See proof below

Explanation:

Important points

  • understanding what it means to be "onto"
  • the nature of a quadratic function
  • finding a value that isn't in the range

Onto

For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.

However, the co-domain here is suggested to be
\mathbb R, whereas the range of f is not
\mathbb R (proof below).

Proof (contradiction)

Suppose that f is onto
\mathbb R.

Consider the output 7 (a specific element of
\mathbb R).

Since f is onto
\mathbb R, there must exist some input from the domain
\mathbb R, "p", such that f(p) = 7.

Substitute and solve to find values for "p".


f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2

Next, apply the square root property:


\pm √(-1) =√(p^2)

By definition,
√(-1) =i, so


i=p \text{ or } -i =p

By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots. Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.

However, neither
i nor
-i are in
\mathbb R, so there are zero values of p in
\mathbb R for which f(p)= 7, which is a contradiction.

Therefore, the contradiction supposition must be false, proving that f is not onto
\mathbb R

User Ngatirauks
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