Question

The function $f\left(x\right)=x^{\frac{1}{3}}(x-1)$

• A. Has one stationary point
• B. Has two critical points
• C. Has two stationary points
• D. Has one point of local minima

Answer 1

Answer: (A), (B), (D)

Has one point of local minima

We have,
$f\left(x\right)=x^{\frac{1}{3}}(x-1)=x^{\frac{4}{3}}-x^{\frac{1}{3}}\Rightarrow f^{\prime }\left(x\right)=\frac{4}{3}{x}^{\frac{1}{3}}-\frac{1}{3}\left(\frac{1}{x^{\frac{2}{3}}}\right)=\frac{1}{3x^{\frac{2}{3}}}[4x-1]$
For stationary points $f^{\prime }\left(x\right)=0$
$\Rightarrow x=\frac{1}{4}$ is the only stationary point
and $x=0$ and $x=\frac{1}{4}$ are the two critical point,
$\because$ at $x=0$ function has a vertical tangent
Now, $f^{\prime }\left(x\right)$ changes its sign from negative to positive as it crosses $x=\frac{1}{4},$ from left to right, which is point of minima.