How? Consider function $f : \mathbb {R} \rightarrow \mathbb {R}$ defined as $f (x)=x^2 + 3$. Find $f ([-3,5])$ and $f^{-1} ([12,19])$.

Question

I have the following question as homework and have figured out the below. I know this is incorrect. How do you calculate these without listing every single iteration of the interval? I'm not looking for the answer, I'm looking for help on how to get to the correct answer.

Consider function $f : \mathbb {R} \rightarrow \mathbb {R}$ defined as $f (x)=x^2 + 3$. Find $f ([-3,5])$ and $f^{-1} ([12,19])$.

I have the following:

$f(x) = x^2 + 3$

domain: $\mathbb {R}$

range: ${f(x)|x ∈ R: f(x) ≥ 3}$

so:

$f(-3) = 12$

$f(5) = 28$

$f([-3,5]) = [12,28]$

$f^{-1}(12)=\pm 148$

$f^{-1}(19)=\pm 364$

$f^{-1}([12,19]) = [-148, 364]$

I know this is a parabola. I also know that my answers are correct for what I have calculated, however, I do not understand how since the function includes 0 between the interval [-3,5] which my interval should be including 3, but it doesn't.

Answer 1

You can't just substitute endpoints to find $f([a,b])$.

A good way to start is by drawing the graph of $f(x)$. Please do this before reading further.

Now $f([-3,5])$ means the set of all values $y=f(x)$, when $x$ takes all values from $-3$ to $5$. So look at your graph, visualise $x$ starting at $-3$ and increasing until it reaches $5$, and think about the corresponding $y$ values. I think you will find it easy to see it starts at $y=12$ and then decreases to $y=3$ (when $x=0$) before increasing again to $y=28$. So $y$ takes all values from $3$ to $28$ (some of them twice, but this doesn't matter), and the answer is $[3,28]$.

For the $f^{-1}$ problem, the first thing to realise is that "$f^{-1}$ of a number" is not at all the same as "$f^{-1}$ of a set". Perhaps the first and most important difference is that $f^{-1}$ of a number might not exist, but $f^{-1}$ of a set always does. In some ways it is very unfortunate that the same notation is used for the two concepts, but that's how it is and we have to live with it.

Anyway... as usual, the first thing is to make sure that you know what the notation means: $f^{-1}(A)$ is the set of all $x$ values such that $f(x)$ is in the set $A$. So in your case we have to solve $$12\le x^2+3\le19\ .$$ If you are up to speed on inequalities you should easily be able to solve this to get $$-4\le x\le-3\quad\hbox{or}\quad 3\le x\le4\ .$$ So the solution is $$\{x\in{\Bbb R}\mid -4\le x\le-3\ \hbox{or}\ 3\le x\le4\}\ ,$$ or if you prefer it in interval notation, $$[-4,-3]\cup[3,4]\ .$$

Answer 2

If a function is increasing and continuous on an interval, then it is true that $$ f((a,b))=(f(a),f(b)) $$ similarly if it is decreasing and continuous, $$ f((a,b))=(f(b),f(a)) $$

On the interval $[-3,5]$, you may break it into pieces; one on which it decreases, and one on which it increases. On $[0,3]$ the function is decreasing, yielding the image $[f(0),f(-3)]=[3,12]$ and then for the next piece, $[0,5]\mapsto [3,28]$. So taking the union of the two sets, we find that the image on this interval is $$ [3,28] $$

As for the preimage of $[12,19]$, this involves solving the inequality $$ 12<x^2+3<19\implies 9<x^2<16\implies 3<|x|<4 $$