0
$\begingroup$

I've come across the following exponential inequality and, unfortunately, I encountered some difficulties trying to solve it.

$$ \left | x \right |^{2x^2 - 3x + 1} \leq 1, x \in \mathbb{R} $$

I tried to solve it by rewriting $ 1 $ as $ \left | x \right | ^ 0 $. After this substitution I got the equivalent inequation $ 2x^2 - 3x + 1 \leq 1 $, which led to $ x \in \left [ \frac{1}{2}, 1 \right ] $. I am not sure about whether this way of solving it is correct or not.

The solution provided by the book approaches two cases, one with $ \left | x \right | \leq 1 $ and the other one with $ \left | x \right | > 1 $. Now, here is where I encounter difficulties. I know how to solve the second case, but I don't really know what I should do in the first case, where $ \left | x \right | \leq 1 $.

So, I think I need some help from you.

Thnak you in advance!

$\endgroup$
  • $\begingroup$ When $|x|<1$, the exponential is not a growing function and the direction of the inequality reverses itself. $\endgroup$ – Yves Daoust Jul 4 '16 at 8:32
1
$\begingroup$

$$ \left | x \right |^{2x^2 - 3x + 1} \leq 1\tag1$$

I tried to solve it by rewriting $ 1 $ as $ \left | x \right | ^ 0 $. After this substitution I got the equivalent inequation $ 2x^2 - 3x + 1 \leq 1 $, which led to $ x \in \left [ \frac{1}{2}, 1 \right ] $. I am not sure about whether this way of solving it is correct or not.

(you should mean $2x^2-3x+1\leq \color{red}{0}$) This is not correct. This is not equivalent to $(1)$ because you assume that $|x|\gt 1$. In other words, this is equivalent to $(1)$ only when $|x|\gt 1$. (by the way, there is no $x$ such that $\frac 12\le x\le 1$ and $|x|\gt 1$.)

The solution provided by the book approaches two cases, one with $ \left | x \right | \leq 1 $ and the other one with $ \left | x \right | > 1 $. Now, here is where I encounter difficulties. I know how to solve the second case, but I don't really know what I should do in the first case, where $ \left | x \right | \leq 1 $.

For $|x|=1$ or $x=0$, the inequality $(1)$ holds.

For $0\lt |x|\lt 1$, note that $(1)\iff 2x^2-3x+1\color{red}{\ge}0$.

The answer is that $\color{red}{-1\le x\le\frac 12\quad\text{or}\quad x=1}$.

$\endgroup$
4
$\begingroup$

hint: Take log we have: $(2x^2-3x+1)\log |x| \leq 0$. You can have two cases here to continue, but simple.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.