Question

Number of critical points on the graph of the function $f\left(x\right)=x^{\frac{1}{3}}(x-4)$ is

• A. 0
• B. 1
• C. 2
• D. None

Answer 1

Answer: (A)

0

Given,

$f\left(x\right)=x^{\frac{1}{3}}(x-4)$

since $f^{\prime }\left(x\right)=0$ as its undefined.

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(x^{\frac{1}{3}}(x-4)\right)$

$=\frac{d}{dx}\left(x^{\frac{1}{3}}\right)(x-4)+\frac{d}{dx}(x-4)x^{\frac{1}{3}}$

$=\frac{1}{3x^{\frac{2}{3}}}(x-4)+1\cdot x^{\frac{1}{3}}$

$=\frac{4x-4}{3x^{\frac{2}{3}}}$

$\therefore \frac{4x-4}{3x^{\frac{2}{3}}}=0$

$\Rightarrow 3x^{\frac{2}{3}}=0\Rightarrow x=0$

Therefore the critical point is: $0$