Question

Prove that the function

Answer 1

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`