Prove that the function

Answer:
See proof below
Step-by-step explanation:
Important points
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 [tex]\mathbb R[/tex], whereas the range of f is not [tex]\mathbb R[/tex] (proof below).
Proof (contradiction)
Suppose that f is onto [tex]\mathbb R[/tex].
Consider the output 7 (a specific element of [tex]\mathbb R[/tex]).
Since f is onto [tex]\mathbb R[/tex], there must exist some input from the domain [tex]\mathbb R[/tex], "p", such that f(p) = 7.
Substitute and solve to find values for "p".
[tex]f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2[/tex]
Next, apply the square root property:
[tex]\pm \sqrt{-1} =\sqrt{p^2}[/tex]
By definition, [tex]\sqrt{-1} =i[/tex], so
[tex]i=p \text{ or } -i =p[/tex]
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 [tex]i[/tex] nor [tex]-i[/tex] are in [tex]\mathbb R[/tex], so there are zero values of p in [tex]\mathbb R[/tex] for which f(p)= 7, which is a contradiction.
Therefore, the contradiction supposition must be false, proving that f is not onto [tex]\mathbb R[/tex]