Answer:
See proof below
Step-by-step explanation:
<u>Important points</u>
- understanding what it means to be "onto"
- the nature of a quadratic function
- finding a value that isn't in the range
<u>Onto</u>
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
, whereas the range of f is not
(proof below).
<u>Proof (contradiction)</u>
Suppose that f is onto
.
Consider the output 7 (a specific element of
).
Since f is onto
, there must exist some input from the domain
, "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](https://tex.z-dn.net/?f=f%28x%29%3D-3x%5E2%2B4%5C%5Cf%28p%29%3D-3%28p%29%5E2%2B4%5C%5C7%3D-3p%5E2%2B4%5C%5C3%3D-3p%5E2%5C%5C-1%3Dp%5E2)
Next, apply the square root property:
![\pm \sqrt{-1} =\sqrt{p^2}](https://tex.z-dn.net/?f=%5Cpm%20%5Csqrt%7B-1%7D%20%3D%5Csqrt%7Bp%5E2%7D)
By definition,
, so
![i=p \text{ or } -i =p](https://tex.z-dn.net/?f=i%3Dp%20%5Ctext%7B%20or%20%7D%20-i%20%3Dp)
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
nor
are in
, so there are zero values of p in
for which f(p)= 7, which is a contradiction.
Therefore, the contradiction supposition must be false, proving that f is not onto ![\mathbb R](https://tex.z-dn.net/?f=%5Cmathbb%20R)