Question-6

Consider the function \(f(x)=x^3-6x\). Which of the following options are correct?

• \(f\) has neither local maxima nor local minima.
• \(\sqrt{2}\) is a local minimum.
• \(\sqrt{2}\) is a local maximum.
• \(-\sqrt{2}\) is a local maximum.
• \(-\sqrt{2}\) is a local minimum.
• \(f\) has two critical points.

Answer

• \(f\) has neither local maxima nor local minima.
• \(\sqrt{2}\) is a local minimum.
• \(\sqrt{2}\) is a local maximum.
• \(-\sqrt{2}\) is a local maximum.
• \(-\sqrt{2}\) is a local minimum.
• \(f\) has two critical points.

Solution

We have:

\[ f( x) =x^{3} -6x \]

The first and second derivatives are:

\[ f^{\prime }( x) =3x^{2} -6,\ f^{\prime \prime} ( x) =6x \]

The critical points of \(\displaystyle f\) are:

\[ \begin{aligned} 3x^{2} -6 & =0\\ x^{2} & =2\\ x & =\pm \sqrt{2} \end{aligned} \]

We have two critical points. Now for the determination of the nature of the optimum:

\[ \begin{aligned} f^{\prime \prime} \left(\sqrt{2}\right) & >0\\ f^{\prime \prime} \left( -\sqrt{2}\right) & < 0 \end{aligned} \]

\(\displaystyle \sqrt{2}\) is a local minimum and \(\displaystyle -\sqrt{2}\) is a local maximum.