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Find the intervals of concavity and the points of inflection of the following functions: $f(x)=2x^{3}+5x^{2}-4x$

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  • Test for concavity Suppose $f$ is twice differentiable on an interval I.
  • (i) If $f''(x) > 0$ for all $x \in I$, then the graph of is concerve upward (convex downward ) on I.
  • (ii) If $ f''(x) < 0$ fro all $x \in I$. then the graph of f is conceve downward ( convex upward) on I.
$f(x)=2x^3+5x^2-4x$
Step 1:
$f'(x)=6x^2+10x-4$
$f''(x)=12 x +10$
$f''(x)=0 => x=\large\frac{-10}{12}=\frac{-5}{6} $$ < 0$ for all values of $x \in R.$
There is no point of inflection and the curve is concave downward every where
answered Aug 5, 2013 by meena.p
 

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