Functions : $ f(x) = {2x-1\over x^2} $

Question

We have :

$$ f(x) = {2x-1\over x^2} $$

1- Determine $ D_f $ and solve the equation $ f(x) = 1 $

2- Show that for every $ x $ from $\mathbb{R}^*_+ $ ; $f(x) \le 1 $

The first exercise is already done and here are my solutions :

$$ D_f = \mathbb{R} - \{0\} $$

By solving $ f(x) = 1 $ I've got $ x = 1 $

Sorry I didn't have much time to write how I did it. The second exercise is my real problem cause I didn't even understand the question to answer it. I did a few drafts :

For every $ x $ from $\mathbb{R}^*_+ $ Could it mean that $ x>0 $ ?

I would imagine that is what it means, although I can't guarantee it not seeing the original textbook/question source. — Chill2Macht, Apr 28 '16 at 22:44
$D_f$ is supposed to be the domain of $f$? I thought you meant the derivative at first — Chill2Macht, Apr 28 '16 at 22:46
You are on the right path. You have to show that for $x > 0$, $f$ has a maximum value of $1$. — N. F. Taussig, Apr 28 '16 at 22:51
The argument provided by egreg demonstrates that the restriction that $x > 0$ is not necessary. As you can check, if $x < 0$, then $f(x) < 0$. Thus, $f$ can only assume positive values when $x > 0$. — N. F. Taussig, Apr 29 '16 at 00:10

score 3 · Answer 1 · answered Apr 28 '16 at 23:09

The domain is correct; the solution to $f(x)=1$ is correct too: indeed $$ \frac{2x-1}{x^2}=1 $$ becomes $$ (x-1)^2=0 $$

Now for the inequality it is essentially the same. You want to see for what value of $x$ you have $f(x)\le1$, that is, $$ \frac{2x-1}{x^2}\le1 $$ This translates into $$ (x-1)^2\ge0 $$ so every $x\in D_f$ satisfies the inequality.

Or you can consider $$ \frac{2x-1}{x^2}=\frac{-x^2+2x-1+x^2}{x^2} =-\frac{(x-1)^2}{x^2}+1 $$ which is clearly $\le1$

score 0 · Answer 2 · edited Apr 29 '16 at 09:54

0

We can first calculate the 1st derivative.

$$\frac{df}{dx} = \frac{d(2x-1)}{dx} \cdot x^{-2} + \frac{d(x^{-2})}{dx} \cdot (2x-1) = \frac{2}{x^2} - \frac{2(2x-1)}{x^3}$$ Set $\frac{df}{dx} = 0$, we have $x =1$. So the turning point is $x =1$.

By calculating the second derivative, we can find out that this point is the maximum of the function. $f(x=1) = 1$. Hence we can conclude for every $x$ from $\mathbb{R}^∗_+$ ; $f(x) \leq 1$.

edited Apr 29 '16 at 09:54

N. F. Taussig

76,571

answered Apr 28 '16 at 23:06

Wes

133

Welcome to MathSE. Judging by her profile, Manal has not yet taken calculus. An observation about your solution: If you simplify the derivative, you obtain $$f'(x) = \frac{2x - 2(2x - 1)}{x^3} = \frac{-2x + 2}{x^3} = -\frac{2(x - 1)}{x^3}$$ so it is easier to use the First Derivative Test in this case. For future reference, you can obtain $\mathbb{R}$ and $\leq$ by typing \mathbb{R} and \leq, respectively, when you are in math mode. Here is a tutorial on how to typeset mathematics on this site. – N. F. Taussig Apr 29 '16 at 10:02

Functions : $ f(x) = {2x-1\over x^2} $

2 Answers2