The real number

Solution:

Let the real number be x.

The cube of the number is $x^{3}$.

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=1-3 x^{2}$

For maxima or minima,

$f^{\prime}(x)=0$

$\Rightarrow 1-3 x^{2}=0$

$\Rightarrow x^{2}=\frac{1}{3}$

$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$

Now,

$f^{\prime \prime}(x)=-6 x$

At $x=-\frac{1}{\sqrt{3}}$, we have

$f^{\prime \prime}\left(-\frac{1}{\sqrt{3}}\right)=-6 \times\left(-\frac{1}{\sqrt{3}}\right)=2 \sqrt{3}>0$

At $x=\frac{1}{\sqrt{3}}$, we have

$f^{\prime \prime}\left(\frac{1}{\sqrt{3}}\right)=-6 \times \frac{1}{\sqrt{3}}=-2 \sqrt{3}<0$

Thus, the real number which must exceeds its cube is $x=\frac{1}{\sqrt{3}}$.

The real number which must exceeds its cube is $\frac{1}{\sqrt{3}}$