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# Find the intervals of concavity and the points of inflection of the following functions: $y=12x^{2}-2x^{3}-x^{4}$

Toolbox:
• 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.
$y=12x^2-2x^3-x^4$
Step 1:
$\large\frac{dy}{dx}$$=24 x-6x^2-4x^3 \large\frac{d^2y}{dx^2}$$=24 -12x-12x^2$
$\qquad=-12(x^2+x-2)$
$\qquad=-12(x+2)(x-1)$
Step 2:
$\large\frac{d^2y}{dx^2}$$=0=>x=-2,1 We consider the intervals (-\infty,-2),(-2,1),(1,\infty) Interval (-\infty,-2) \large\frac{d^2y}{dx^2}$$=(-)(-)(-)=(-)$
Concavity downward
Interval $(-2,1)$
$\large\frac{d^2y}{dx^2}$$=(-)(+)(-)=(+) Concavity upward Interval (1,\infty) \large\frac{d^2y}{dx^2}$$=(-)(+)(+)=(-)$
Concavity downward
The curve changes from concave downward to concave upward when $x=-2$
The corresponding value of $y=12(4)-2(-8)-16$
$\qquad= 48+16-16$
The point of inflection is $(-2,48)$
The curve changes form concave upward to concave downward at $x=1$ .
The corresponding value of $y=12(1)-2(1)-1=9$
The point of inflection is $(1,9)$
edited Apr 29, 2014 by meena.p