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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 $
AMUAMU 2010Application of Derivatives
Solution:
$f \left(x\right)=\frac{4}{3} x^{3}-8x^{2}+16x+5 \ldots\left(i\right)$
There will be a point of inflexion at point $x$
if $\left(\frac{d^{2}y}{dx^{2}}\right)_{at \left(x=2\right)} =0$, but $\frac{d^{3}y}{dx^{3}} \ne0 $
From Eq. $\left(i\right)$, $f'\left(x\right)=4x^{2}-16x+16$
$f''\left(x\right)=8x-16$
$f''\left(2\right)=16-16=0$
and $f'' \left(x\right)=8\ne0$
Hence, at $x = 2, f\left(x \right)$ shown point of inflexion