Injectivity and surjectivity of a funcion

Hi,

can anyone explain me how to calculate surjectivity and injectivity of a function on a following example

f(x) = -x^2 + 3

I know that the function is infective if f(x1) = F(x2) = -x^2+3 = -x^2+3 so -x^2 = -x^2 (hope that's correct)

and if I take f(3) = -6 and f(-3) = -6

so the function is not injective

for surjectivity y = -x^2+3 but I don't know why it is NOT surjective...

Is there any simpler way to go through this. I don't know how I should show the answers on the exam...

Re: Injectivity and surjectivity of a funcion

Yes, a function, f, is "injective" if and only if $\displaystyle f(x_1)= f(x_2)$ only when $\displaystyle x_1= x_2$. You are correct that, for this example, $\displaystyle f(x_1)= f(x_2)$ means that $\displaystyle -x_1^2+3= -x_2^2+ 3$ which leads to $\displaystyle x_1^2= x_2^2$ and then **either** $\displaystyle x_1= x_2$ **or** $\displaystyle x_1= -x_2$. Because of the second possibility, this function is not injective.

As for "surjective", that means that for any y, there exist x such that f(x)= y. To show that this function is NOT surjective, it is enough to demonstrate a number, y such that there is NO x that makes $\displaystyle y= -x^2+ 3$. And one way to do **that** is to note that $\displaystyle x^2$ is never negative so that $\displaystyle -x^2$ is never positive and, finally, that $\displaystyle f(x)= -x^2+ 3$ is **never** larger than 3. A very simple way of showing that this function is not surjective is pick y larger than 3, 4, say, and show that $\displaystyle f(x)= -x^2+ 3= 4$ has no solution.

The very simplest way of seeing that this function is neither injective nor surjective is to observe that its graph is a parabola, opening downard, with vertex at (0, 3). Do you see why that immediately show the function is neither injective nor surjective?

Re: Injectivity and surjectivity of a funcion

ok I see, but where did you take x1 = -x2 from? and also can I calculate surjectivity from preimage of a function?

say f^-1(4) = -4^2 - 3 = 13 but then I dont know how to make a conclusion...

Re: Injectivity and surjectivity of a funcion

I've never been good at this stuff. However I would like to ask...if the domain and range of a function are not given then are we to assume at the outset that the function is $\displaystyle f: \mathbb{R} \to \mathbb{R}$? My thought is that every time I've done a problem like this at least a domain is given. And there is a way to define f for which y = -x^2 + 3 is actually a bijection. For example: $\displaystyle f: \[0, \sqrt{3} \] \rightarrow \[ 0,3 \] $

-Dan

Re: Injectivity and surjectivity of a funcion

how did you get [0,3]? and there shouldnt be minus square root of 3?

and also, is the square function always NOT injective? because there always will be 2 solutions.

Thx

Re: Injectivity and surjectivity of a funcion

f is symmetric about the y-axis (you get the same function if you replace x with -x) so if we include an interval with both a and -a f will not be injective.

to ensure we get surjectvity, we need to restrict the "target set" (the allowable range) to (at most) the image of our restricted domain. f(x) = -x^{2} + 3 is injective on [0,√3], which means we can use the endpoints of [0,√3] to determine appropriate limits of our range:

f(0) = -0^{2} + 3 = 3

f(√3) = -(√3)^{2} + 3 = -3 + 3 = 0

so on the (restricted) domain [0,√3], f only takes values (and takes ALL values, because of something known as the "intermediate value theorem") in [0,3]

the squaring function is NOT "always NOT injective". for example, if we insist that we are only considering x ≥ 0, then there is no question of "which square root to take". likewise to ensure the squaring function is surjective, we must not allow negative numbers in the range, for no real number squared is ever negative.

Re: Injectivity and surjectivity of a funcion

I still don't understand why that function is not surjective and how should I calculate it. Could you explain it in a simpler way please?