Question

If $f\left(x\right)=\frac{x^3}{3}-x^2+\frac{4}{3}$ has maximum value ${}^{\prime }{a}^{\prime }$ at $x=b,$ then $(a,b)$ is

• A. $(\frac{4}{3},0)$
• B. $(2,-\frac{4}{3})$
• C. $(2,\frac{4}{3})$
• D. $(0,\frac{4}{3})$

Answer 1

The correct option is A $(\frac{4}{3},0)$

Given, $f\left(x\right)=\frac{x^3}{3}-x^2+\frac{4}{3}$

$\Rightarrow f^{\prime }\left(x\right)=x^2-2x=0$
$\Rightarrow x=0,2$

Now, $f^{\prime \prime }\left(x\right)=2x-2$
$\Rightarrow f^{\prime \prime }\left(2\right)=2>0$
and $f^{\prime \prime }\left(0\right)=-2<0$
$\therefore f\left(x\right)$ has maximum at $x=0$ which equals to $f\left(0\right)=\frac{4}{3}$