A) local maxima
B) local minima
C) point of inflection
D) none of these
Correct Answer: C
Solution :
[c] : \[f(x)=\frac{4}{3}{{x}^{3}}-8{{x}^{2}}+16x+5\] ...(i) Differentiating (i) with respect to x, we get \[f'(x)=\frac{4}{3}\times 3{{x}^{2}}-16x+16=4{{x}^{2}}-16x+16\] Now for maximum/minimum we put\[f'(x)=0\] \[\Rightarrow \]\[{{x}^{2}}-4x+4=0\Rightarrow {{(x-2)}^{2}}=0\Rightarrow x=2\] \[f''(x)=8x-16,f''(x){{|}_{at\,x=2}}=0\] \[f'''(x)=8\ne 0\] \[\therefore \] x = 2 is the point of inflection,
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