3
$\begingroup$

$$ 2^x > 1-x, \\ e^{x\ln2} > 1-x, \\ (1-x)e^{-x\ln2} < 1, \\ (1-x)e^{(1-x)\ln2} < e^{\ln 2} = 2^1, \\ (1-x)^2e^{\ln2} < e^{\ln 2}, \\ (1-x)^2 < 1, \\ x^2-2x+1-1 < 0, \\ x(x-2) < 0. \\ $$ and then I did: $$ x(x-2) = 0, \\ x_1=0 , x_2=2 \\ $$ The solution is: $0<x<2 $ or $ x∈(0,2). \\$

I would like to know if my way of solving this inequallity is correct? (I do not have the solutions, so I do not know if $0<x<2 $ is the solution of this inequallity.)

Is there an easier way of solving this? Besides solving it graphically.

Thank you in advance!

$\endgroup$
11
  • 1
    $\begingroup$ Have you checked your solution? Does the inequality hold when x = -3, x = 1, x = 5? $\endgroup$
    – Improve
    May 24 '14 at 11:37
  • $\begingroup$ @Improve: How do you mean? I do not have the solutions, so I do not know if it is correct. $\endgroup$ May 24 '14 at 11:39
  • $\begingroup$ Your third step does not make sense to me - what were you trying to do there? You can see that your answer is not correct though. For $x > 0$, $2^x$ gets bigger as $x$ increases and $1-x$ gets smaller as $x$ increases. Since $2^{x} > 1- x$ for $x > 0$, why should your interval of validity be bounded above by $2$? Try focusing on $x \leq 0$. $\endgroup$ May 24 '14 at 11:40
  • $\begingroup$ Well. Is for instance $2^5 > 1 - 5$? In that case, $5$ should be one of your solutions. $\endgroup$
    – Improve
    May 24 '14 at 11:40
  • 1
    $\begingroup$ Part of your question was if the solution is correct. It is important that you are to some extent able to verify or disprove this. Are you sure you know what it means for a value to be a solution of your given inequailty? Your proposed solution have you an interval of $ x \in (0,2)$, which I must say is clearly wrong if you test some values of x. The left-hand side is positive for all x, while the right-hand side is negative for all $x > 1$, so the inequality at least holds for all $x>1$. Whenever you solve an inequality, you should ask yourself: Does my solution make sense? $\endgroup$
    – Improve
    May 24 '14 at 11:49
3
$\begingroup$

$$f(x)=2^x-1+x\Rightarrow f'(x)=(e^{\ln 2^x})'+1=e^{x\ln 2}\ln2+1>0$$ this tell us that this function is strictly increasing in the whole line. Since $f(0)=0$ then you require inequality is satisfied by $x>0$.

$\endgroup$
3
$\begingroup$

The left side is everywhere strictly increasing, and the right side is everywhere strictly decreasing; so the two sides are either equal for exactly one value of $x$ or not at all. By inspection, they are equal at $x=0$, so "$<$" holds for $x<0$ and "$>$" holds for $x>0$. Don't underestimate the power of graphical methods!

This is really equivalent to the answer posted by Luis Valerin.

$\endgroup$
1
  • 1
    $\begingroup$ But i like more your answer+1. $\endgroup$
    – Valent
    May 24 '14 at 12:03

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.