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Find the intervals of concavity and the points of inflection of the following functions: $f(x)=x^{4}-6x^{2}$

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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.
Step 1:
$f(x)=x^4-6x^2$
$f'(x)=4x^3-12x$
$f''(x) =12 x^2-12$
Step 2:
$f''(x)=0=>12x^2-12=>(x+1)(x-1)=0$ or $x=-1,1$
We consider the intervals $ (-\infty,-1),(-1,1),(1,\infty)$
Interval $(-\infty,-1) $
$f''(x)=12(x+1)(x-1)=>-$
Concavity downward
Interval $(-1,1) $
$f''(x)=12(x+1)(x-1)=>+$
Concavity upward
Interval $(1,\infty) $
$f''(x)=12(x+1)(x-1)=>-$
Concavity downward
Step 3:
The curve changes from concave downward to concave upward when $x=-1$
$f(-1)=1-6=-5$
$\therefore (-1,5)$ is a point of inflection
The curve changes from concave upward to concave downward at $x=1$ where $f(x)= -3$
$\therefore (1,-5)$ is also a point of inflection
answered Aug 5, 2013 by meena.p
 

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