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Q.
For the function $f(x)=\frac{4}{3}x^3-8x^2+16x+5,\, x=2$ is a point of
Application of Derivatives
Solution:
$f\left(x\right)=\frac{4}{3}x^{3}-8x^{2}+16x+5\quad\ldots\left(1\right)$
Differentiating with respect to $x$, we get
$f'\left(x\right)=\frac{4}{3} \times 3x^{3}-3x^{3}-16x+16$
Now for maximum/minimum we put $f'\left(x\right) = 0$
$\Rightarrow x^{2}-4x+4=0$
$\Rightarrow \left(x-2\right)^{2}=0$
$\Rightarrow x=2$
$f''\left(x\right)=8x-16$,
$f''\left(x\right)|_{x=2} =0$
$f'''(x)=8 \ne 0$
$\therefore x=2$ is the point of inflexion