$x\sqrt{x-1}$ bijectivity

Question

So I know to prove that in order to prove the function is a bi-jectivity I have to solve the equation $x\sqrt{x-1}= y$ and find one and only one solution, and I am having trouble doing it. The function is from $[1 ; +\infty[ \to [0 ; +\infty[$

Answer 1

Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:

Let $f(x)= x \sqrt{x-1}$.

Surjectivity: First, what is $f(1)$? What is $\lim_{x \to \infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $\infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]

Injectivity: Find $f'(x)$? Given your domain is $[1,\infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:

Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]

Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,\infty)$ to $[0,\infty)$.

Answer 2

1) Injective.

$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.

Let $1 \le x_1 < x_2.$

$g(x):=√x$ is stricly increasing.

For $a,b \ge 0:$

$a-b=(√a-√b)(√a+√b).$

$a-b >0$ implies $√a-√b >0$ (why?).

Hence

$(x_1-1)^{1/2} \lt (x_2-1)^{1/2}$, and

$x_1(x_1)^{1/2} \lt x_2(x_2-1)^{1/2}$(why?), i.e

strictly increasing, injective.

2) Bijective

Let $y > 0 =f(0). $

Since $\lim_{x \rightarrow \infty}f(x)= \infty$ , $f$ is not bounded above . There is $b>0$ s.t.

$y < f(b)$.

Consider the continuos function $f$ on

$[0,b]$, with $f(0)\lt y \lt f(b)$.

Intermediate Value Theorem(Corollary):

There is a $p \in [0,b]$ with $f(p)=y$,

hence surjective.