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How do you solve :

$$2^x < x^2$$

My math years are behind me, so I can't wrap my head around how to continue after this step :

$$2^x - x^2 < 0$$

I think there's a trick since it's a 0 comparison inequality, but since the 2 parts have different powers, I'm not so sure...

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  • 2
    $\begingroup$ There is no trick here. I suggest that you start by drawing the two graphs on the same set of axes. That might suggest an argument. $\endgroup$ – Ethan Bolker Apr 28 '16 at 14:49
  • $\begingroup$ $\ln$ both sides. $\endgroup$ – Kenny Lau Apr 28 '16 at 14:51
  • $\begingroup$ But it is slightly tricky if you allow $x$ to be any real value. There are three values of $x$ where they are equal. $\endgroup$ – almagest Apr 28 '16 at 14:52
  • $\begingroup$ @N.S.JOHN But $2^x=x^2$ for $x=4$. $\endgroup$ – almagest Apr 28 '16 at 14:53
  • $\begingroup$ @almagest you are right. $\endgroup$ – N.S.JOHN Apr 28 '16 at 14:54
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The function $f(x):=2^x-x^2$ has zeros at $x=2$ and $x=4$. Furthermore from $f(0)=1$ and $f(-1)=-{1\over2}$ it follows that there is a zero $\xi\in\ ]{-1},0[\ $ where $f$ changes sign. On the other hand, one computes $$f''(x)=(\log2)^2 \cdot 2^x-2\ .$$ Since $\log2\doteq0.693$ it follows that $f''$ is monotonically increasing from negative to positive values; hence $f''$ has exactly one zero. Therefore by Rolle's theorem $f'$ has at most two zeros, hence $f$ has at most three zeros. These three zeros have already been identified; furthermore one easily verifies that $f'(2)\ne0$, $f'(4)\ne0$. This shows that $f$ changes sign at each of its zeros, and as $f$ is certainly $>0$ for large $x$ we can say that $f$ is negative for $2<x<4$ and for $x<\xi\doteq-{23\over30}$.

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