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I have just started my course in calculus and would need some guidance here. If possible, don't give complete solutions, since I need all training I can possibly get :)

Problem. The function $f$ is defined by $$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}\,,\:\: x>0\,.$$ Show that $f$ is injective in as many ways as possible.

My thoughts.

I have started by showing that $f$ is strictly increasing by showing that $f(x+\epsilon)-f(x)>0$ for any positive $\epsilon$ (we have not introduced derivatives yet, so I guess that was the best way to do it, but perhaps there are more efficient methods?). By the definition of injectivity, that implies that $f$ also is injective. Is that an acceptable solution?

I also tried to find an inverse by solving $y=f(x)$ for $x$. This gave me $$\sqrt{x}=\frac{1-y}{1+y}\Longrightarrow x=\left(\frac{1-y}{1+y}\right)^2,$$ but I am not entirely sure what I should do about the implication. Actually, I am not even sure if it's a problem.

Except this, I also wonder if there are any other ways that I can show the injectivity of $f$.

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    $\begingroup$ Your approach is fine. Now suppose that there are $x_1$ and $x_2$ so that $f(x_1)=f(x_2)=y$. What does your equation say about $x_1$ and $x_2$? $\endgroup$ – robjohn Sep 15 '13 at 22:42
  • $\begingroup$ It says that $$f(x_1)=f(x_2)=y\Longrightarrow x_1=x_2=\left(\frac{1-y}{1+y}\right),$$ which is logically equivalent with the definition of injectivity, $$x_1\neq x_2\Longrightarrow f(x_1)\neq f(x_2).$$ So, I suppose that means that I don't need to show that $$x=\left(\frac{1-y}{1+y}\right)\Longrightarrow \sqrt{x}=\frac{1-y}{1+y},$$ right? (But if I wanted to, how would I do it?) $\endgroup$ – Oskar Henriksson Sep 15 '13 at 23:15
  • $\begingroup$ Note that your function is decreasing. $\endgroup$ – André Nicolas Sep 15 '13 at 23:18
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To show injectivity, you need to show that if $f(x_1)=f(x_2)$, then $x_1=x_2$. Your approach works just fine. Suppose that $f(x_1)=f(x_2)=y$. Your equation says that $$ x_1=\left(\frac{1-y}{1+y}\right)^2 $$ and $$ x_2=\left(\frac{1-y}{1+y}\right)^2 $$ That is, $x_1=x_2$.

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You can write $f(x)$ as follows: $$ f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}=\frac{2}{1+\sqrt{x}}-1\,,\:\: x>0\,. $$ Then if you accept that $g(x)=\sqrt x$ is injective then $f(x)$ should be injective too.

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To show that $f(x)$ is strictly decreasing, just note that $1-\sqrt{x}$ is strictly decreasing and $1+\sqrt{x}$ is strictly increasing.

Then prove that if $a(x)$ is strictly decreasing and $b(x)$ is strictly increasing and positive, then $\dfrac{a(x)}{b(x)}$ is strictly decreasing.

We want to show that $\dfrac{a(x+h)}{b(x+h)} < \dfrac{a(x)}{b(x)} $.

$\begin{align} \frac{a(x+h)}{b(x+h)} &< \frac{a(x)}{b(x+h)} \quad \text{ (since } a(x+h) < a(x))\\ &< \frac{a(x)}{b(x)} \quad \text{ (since } b(x+h) > b(x) > 0)\\ \end{align} $

That's all you need.

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  • $\begingroup$ Okay. I understand how $$a(x+h)<a(x)\Longleftrightarrow\frac{a(x+h)}{b(x+h)}<\frac{a(x)}{b(x+h)},$$ but I can't convince myself that $$b(x+h)>b(x)>0 \Longleftrightarrow \frac{a(x)}{b(x+h)}<\frac{a(x)}{b(x)}$$ for $a(x)\leq 0$. $\endgroup$ – Oskar Henriksson Sep 16 '13 at 12:27

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