Question

The real number which must exceeds its cube is _______________.

Answer 1

Let the real number be x.

The cube of the number is x³.

Let f(x) = x − x³

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=1-3x^2$

For maxima or minima,

$f\text{'}\left(x\right)=0$

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

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

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

Now,

$f\text{'}\text{'}\left(x\right)=-6x$

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

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

So, $x=-\frac{1}{\sqrt{3}}$ is the point of local minimum of f(x).

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

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

So, $x=\frac{1}{\sqrt{3}}$ is the point of local maximum of f(x).

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


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