State whether the functions are injective, surjective, or bijective.

Question

In the book titled: Discrete Mathematical Structures, by Tremblay; it is stated as Q.#1 in Ex. 2-4.3:

Let $f:\mathbb{R}\to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$, where $\mathbb{R}$ is the set of real numbers. Find $f\circ g$ and $g\circ f$, where $f(x)= x^2 -2$ and $g(x) = x+4$. State whether these functions are injective, surjective, or bijective.

$f\circ g = x^2 +8x +14$,
$g\circ f=x^2 +2$

By my earlier post, have gathered that as an even order function will not be having a unique x-value for each y-value so none is injective.
But, am still not sure about disproving surjectivity. My approach is not based on calculus (that am still unable to understand, as stated on mse here); but is based on disproof by taking cases where failure occurs.

The minimum value taken up by $f\circ g = x^2 +8x +14$ is $14$.
Similarly, the minimum value taken up by $g\circ f = x^2 +2$ is $2$.
Hence, the entire range is not mapped.

Have a further request to make me understand by the calculus approach.

---

Update In face of comment by @Arthur, the correct minimum value for $f\circ g = x^2 +8x +14$ is given for $x=-4$, i.e. $-2$.
This is found by simply differentiating wrt $x$ & equating to $0$ to find the minima point.

Answer 1

Consider a continous function $h$ on $\mathbb{R}$, satisfying $\lim_{x \to \infty} h(x) = \infty$ and $\lim_{x \to -\infty}h(x) = \infty$.

then there exists an $M>0$, such that $|x| > M$, we have $f(x) > 1$.

Also, on $[-M, M]$ which is a compact set, the continous function attains a minimum value $m$.

Hence, it is impossible to attain $\min\{m,1\}-1$, hence it can't be surjective.

If a function has attains a global optimum, it can't be surjective as we can't have values that perform better than the global optimum.

Update:

For example, given any quadratic equation with leading positive coefficient, then it satisfies the condition. I know such $M$exists, but we are not interested in its exact value, I just know it exists. For the example of $f(x)=x^2-2$, I can pick $M$ to be $2$.

Then if you look at any values $x \in (-\infty, -2) \cup (2, \infty)$, we know that $f(x) > 1$.

Now, let's look at $[-2,2]$, we found a minimal value, which is $(0,-2)$.

We can't attain value like $-3$ because we know that on the interval $[-2,2]$, the minimal value is $-2$ and outside $[-2,2]$, their values are bigger than $1$.

Answer 2

We can refer to the general fact that

• $f\circ g:A\to B$ is surjective $\implies f:A\to B$ is surjective

and in that case since $f$ is not surjective then $f\circ g$ can’t be surjective.

We can also note that $g\circ f$ is continuous and

$$ \lim_{x\to \pm \infty} g\circ f=+\infty$$

therefore $g\circ f:\mathbb{R}\to \mathbb{R}$ is bounded from below and it can't be surjective.

As an alternative to the frst method, a similar argument can be used to show that $f\circ g:\mathbb{R}\to \mathbb{R}$ is not surjective.

Answer 3

So, you have sucessfully found $f\circ g$ and $g\circ f$. That's good. Now we want to study these two functions, and see whether they are surjective.

That $f\circ g$ is surjective means that for any real number $r$, we can solve $f\circ g(x) = r$. Specifically, if it is surjective, then we should be able to solve $f\circ g(x) = -10$. So we try that: $$ x^2 + 8x + 14 = -10\\ x^2 + 8x + 24 = 0 $$ At this point we have a standard quadratic equation. If you try solving it, you will see that it cannot be done; the equation has no (real) solutions. So we cannot solve $f\circ g(x) = -10$. This means that we cannot solve $f\circ g(x) = r$ for all real numbers $r$ (as we have just found a counterexample). This again means that $f\circ g$ is not surjective.

You can show that $g\circ f$ is not surjective in exactly the same way. In general, when trying to show that a function is not surjective, this is a popular strategy. However, using the specific number $-10$ clearly doesn't always work. You will have to analyse each function and find a suitable $r$ to use in each case.

Answer 4

composite function: \begin{align*} (f \circ g)(x) & = f(g(x))\\ & = f(x + 4)\\ & = (x + 4)^2 - 2\\ & = x^2 + 8x + 16 -2\\ & = x^2 + 8x + 14 \end{align*}

For the purposes of determining whether the function is injective or surjective, it is helpful to consider the vertex form $$(f \circ g)(x) = (x + 4)^2 - 2$$ By setting the term inside the parentheses equal to zero, we find that the vertex of the parabola is $(-4, - 2)$.

injective: Is $f \circ g: \mathbb{R} \to \mathbb{R}$ defined by $f \circ g)(x) = (x + 4)^2 - 2$ injective?

If $f \circ g$ is injective, then $(g \circ f)(x_1) = (g \circ f)(x_2) \implies x_1 = x_2$. However, we can use symmetry to prove that is not the case here. $$(f \circ g)(-3) = (-3 + 4)^2 - 2 = 1^2 - 2 = 1 - 2 = (-1)^2 - 2 = (-5 + 4)^2 - 2 = (f \circ g)(-5)$$

Since $(f \circ g)(-3) = (f \circ g)(-5)$, $f \circ g$ is not injective.

surjective: Is $f \circ g: \mathbb{R} \to \mathbb{R}$ defined by $(f \circ g)(x) = (x + 4)^2 - 2$ surjective?

For a function to be surjective, its range must equal its codmain, which in this case is $\mathbb{R}$.

Since $(f \circ g)(x) = (x + 4)^2 - 2 \geq -2$, the range of $f \circ g$ is $\text{Ran}_{f \circ g}(x) = [-2, \infty)$. If $f \circ g$ were surjective, then it range would be $\mathbb{R}$. Since $-3$ is not in its range, $f \circ g$ is not surjective.

bijective: Is $f \circ g: \mathbb{R} \to \mathbb{R}$ defined by $(f \circ g)(x) = (x + 4)^2 - 2$ bijective?

Since a function must be both injective and surjective to be bijective, $f \circ g$ is not bijective.

composite function: \begin{align*} (g \circ f)(x) & = g(f(x))\\ & = g(x^2 - 2)\\ & = x^2 - 2 + 4\\ & = x^2 + 2 \end{align*}

Notice that $(g \circ f)(x) = x^2 + 2$ is in vertex form and has vertex $(0, 2)$.

injective: Is $g \circ f: \mathbb{R} \to \mathbb{R}$ defined by $(g \circ f)(x) = x^2 + 2$ injective?

$(g \circ f)(-1) = (-1)^2 + 2 = 1 + 2 = 1^2 + 2 = (g \circ f)(1)$, so $g \circ f$ is not injective.

surjective: Is $g \circ f: \mathbb{R} \to \mathbb{R}$ defined by $(g \circ f)(x) = x^2 + 2$ surjective?

Since $(g \circ f)(x) = x^2 + 2 \geq 2$, the range of $g \circ f$ is $\text{Ran}_f = [2, \infty)$. Since $0$ is not in its range, the function $g \circ f$ is not surjective.

bijective: Is $g \circ f: \mathbb{R} \to \mathbb{R}$ defined by $(g \circ f)(x) = x^2 + 2$ bijective?

Since $g \circ f$ is neither injective nor surjective, it is not bijective.