Question

The function $f(x)=x^{1/3}(x-1)$

• A. has 2 inflection points.
• B. is strictly increasing for x > 1/4 and strictly decreasing for x < 1/4.
• C. is concave down in (- 1/2, 0).
• D. Area enclosed by the curve lying in the fourth quadrant is 9 28

Answer 1

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

$y=x^{1/3}(x-1)$
$\frac{dy}{dx}=\frac{4}{3}{x}^{1/3}-\frac{1}{3}\cdot \frac{1}{x^{2/3}}=\frac{1}{3x^{2/3}}[4x-1]$

$\text{ now }{f}^{\prime }(x)=\frac{4}{3}{x}^{1/3}-\frac{1}{3}\cdot x^{-2/3}\text{ (non existent at }x=0\text{, vertical tangent) }$
$f^{\prime \prime }(x)=\frac{4}{9}\cdot \frac{1}{x^{2/3}}+\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{1}{x^{5/3}}$
$=\frac{2}{9x^{2/3}}\left[2+\frac{1}{x}\right]=\frac{2}{9x^{2/3}}\left[\frac{2x+1}{x}\right]$
$\therefore f^{\prime \prime }(x)=0\text{ at }x=-\frac{1}{2}\text{ (inflection point) }$
$\text{ graph of }f(x)\text{ is as }$

$A{=\underset{0}{\overset{1}{\int }}\left(x^{4/3}-x^{1/3}\right)dx=\frac{3}{7}{x}^{3/7}-\frac{3}{4}{x}^{4/3}]}_0^1$
$=\left|\frac{3}{7}-\frac{3}{4}\right|=3\left|\frac{4-7}{28}\right|=\frac{9}{28}\Rightarrow \text{ (D) }$