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Let $f:(−1,\infty)\to (−1,\infty)$ be defined by $f(x)=x^2+2x $, study the injection and surjection of $f$ , then find the inverse function if exist .

So i showed that the function is not $1-1$. my problem is am struggling with showing whether it’s surjective or not i know it’s surjective if the range of the function =the codomain of the function but i dont know how to .. also the inverse doesn’t exist since it’s not one-to-one ?? any help would be appreciated!

lulu
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user749680
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4 Answers4

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If we think of as a function from $\mathbb R$ to $\mathbb R$ it is neither injective nor surjective.

But, over a restricted domain and co-domain, it could be. We have to find that domain and co-domain. Since $f$ is continuous, it can only be 1-1 if is monotonic. Where is $f$ strictly increasing, or strictly decreasing. Either one of these will give us a suitable restriction for the domain.

And what is the max and min over this restricted domain? This will give us the co-domain.

Now you can look for a suitable inverse.

Doug M
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To understand the function, complete the squares to write $f(x)=(x+1)^2-1$.

Now try to solve $(x+1)^2-1=c$ for some $c\in (-1, \infty)$.

We see that this has real solutions so long as $c≥-1$ and that if $c>-1$ it has two distinct solutions, given by $x=-1\pm \sqrt {1+c}$. Since $\sqrt {1+c}>0$ we see that exactly one of these solutions is $>-1$, hence we get both surjectivity and injectivity and we have found an explicit inverse, namely $$f^{-1}(c)=-1+\sqrt {1+c}$$

lulu
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  • Am so sorry.. i feel so bad that I didn’t understand it – user749680 Feb 11 '20 at 00:19
  • Ok. Try this: If $y>x>-1$ then show that $(y+1)^2>(x+1)^2$ and deduce injectivity that way. – lulu Feb 11 '20 at 00:21
  • But, you must have some examples to work from. Can you prove that the function $x^2$ is injective for $x>0$? This is the same thing! we just translate the function around a little. – lulu Feb 11 '20 at 00:22
  • so u mean i should add and square each side of the inequality x>y>1 to get it to the form (y+1)^2>(x+1)^2 – user749680 Feb 11 '20 at 00:25
  • i have worked on f(x)=5x+2 on the set R – user749680 Feb 11 '20 at 00:27
  • Well, you have to get the signs right. Work the easier example. Show that $x^2$ is injective for $x>0$. – lulu Feb 11 '20 at 00:27
  • So, from that example you should see that constructing a well defined inverse essentially solves the problem. That's what I did, I constructed an inverse that is defined for $c>-1$. – lulu Feb 11 '20 at 00:28
  • f(x)=x^2 for x>0 is injective – user749680 Feb 11 '20 at 00:30
  • Yes, it is. But you should convince yourself that you know how to prove that. Once you have done that, you should see that $(x+1)^2-1$ is, essentially , the same function, just translated. – lulu Feb 11 '20 at 00:34
  • can u please just explain to me your approach, i dont just want the solution i really just want to understand it – user749680 Feb 11 '20 at 00:37
  • I really don't know what I can add to what's been written already. Please study my posted solution, as well as the other posted solution here. If your text has examples, study those as well! – lulu Feb 11 '20 at 00:42
  • ok thank u very much – user749680 Feb 11 '20 at 00:44
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For completeness, I'll prove that the function is bijective.

Injective: Note that for $x_1,x_2\in(-1,\infty)$, if $x_1^2+2x_1=x_2^2+2x_2$, then $x_1^2-x_2^2+2x_1-2x_2=0$, so $(x_1-x_2)(x_1+x_2+2)=0$. Assuming $x_1,x_2$ are distinct, $x_1-x_2\neq0$, so this implies that $x_1+x_2+2=0$, or in other words, $x_1+x_2=-2$. Now, since $x_1,x_2>-1$, $x_1+x_2>-2$, so this is a contradiction, showing injectivity.

Surjective: For any $y>-1$, consider $x=-1+\sqrt{1+y}$. Since $1+y>0$, $\sqrt{1+y}>0$, so we know that $x\in(-1,\infty)$. Moreover, $$(-1+\sqrt{1+y})^2+2(-1+\sqrt{1+y})=1-2\sqrt{1+y}+1+y-2+2\sqrt{1+y}=y$$

Rushabh Mehta
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The function $f(x) = x^2 +2x$ is a bijection if and only if there exists a function $g: (-1, +\infty) \to (-1, +\infty)$ satisfying

$\tag 1 f \circ g = g \circ f = \text{Id}_{(-1, +\infty)}$

Solve for $x$ in terms of $y$:

$\quad y = x^2 + 2x \; \text{ iff } \; x^2 + 2x - y = 0 $

Using the quadratic formula,

$\quad \displaystyle x = \frac{-2 + \sqrt{4 + 4y}}{2} \text{ or } x = \frac{-2 - \sqrt{4 + 4y}}{2} $

Define $g(x): (-1, +\infty) \to (-1, +\infty)$ with

$\tag 2 g(x) = \displaystyle \sqrt{x + 1} - 1$

checking that $g$ is indeed a transformational mapping of the interval $(-1, +\infty)$.

The function $g$ is a left inverse:

$\quad g \big(f (x) \big) = \displaystyle \sqrt{(x^2 + 2x) + 1} - 1 = \sqrt{ (x+1)^2} - 1 = x$

The function $g$ is a right inverse:

$\quad f \big( g (x) \big) = \displaystyle (\sqrt{x + 1} - 1)^2 +2(\sqrt{x + 1} - 1) = (x+1) - 2\sqrt{x + 1} + 1 + 2\sqrt{x + 1} - 2 = x$

There is nothing left to do.

CopyPasteIt
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