Question

Prove that the function

Answer 1

Answer:

See proof below

Step-by-step 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:

$\pm \sqrt{-1} =\sqrt{p^2}$

By definition, $\sqrt{-1} =i$, 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 $\mathbb R$