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

$\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}{9 x ^{2 / 3}}\left[2+\frac{1}{ x }\right]=\frac{2}{9 x ^{2 / 3}}\left[\frac{2 x +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 \left.=\int\limits_0^1\left(x^{4 / 3}-x^{1 / 3}\right) d x=\frac{3}{7} x^{3 / 7}-\frac{3}{4} x^{4 / 3}\right]_0^1 $
$=\left|\frac{3}{7}-\frac{3}{4}\right|=3\left|\frac{4-7}{28}\right|=\frac{9}{28} \Rightarrow \text { (D) }$