How do you find all points of inflection given $y=x^4-3x^2$ ?

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Answer 1 · 2017-08-10T05:51:42

Inflection points at $x = +- 1/sqrt(2)$ .

Find the second derivative.

$y' = 4x^3 - 6x$

$y'' = 12x^2 - 6$

Inflection points occur when $y'' = 0$ .

$0 = 12x^2 - 6$

$0 = 6(2x^2 - 1)$

$x = +- 1/sqrt(2)$

If we select test points, we see that the sign of the second derivative does indeed change at $x = -1/sqrt(2)$ and $x = +1/sqrt(2)$ .

Therefore, $x = -1/sqrt(2)$ and $x = + 1/sqrt(2)$ are both inflection points.

Hopefully this helps!

Answer 2 · 2017-08-10T06:07:32

Point of infection

$(0.7, -1.23)$
$(-0.7, -1.23)$

Given -

$y=x^4-3x^2$

$dy/dx=4x^3-6x$

$(d^2y)/(dx^2)=12x^2-6$

$(d^2y)/(dx^2)=0=>12x^2-6=0$

To find the point of inflection, set the second derivative equal to zero

$x^2=6/12=1/2=0.5$
$x=+-sqrt0.5=0.7$

At $x=0.7$
At $x=-0.7$

$y=0.7^4-3(0.7)^2$

$y=0.2401-1.47=-1.23$
$y=-1.23$

Point of infection

$(0.7, -1.23)$

$y=0.2401-1.47=-1.23$
$y=-0.5096$
$y=-1.23$
Point of infection
$(-0.7, -1.23)$

How do you find all points of inflection given y=x^4-3x^2y=x^4-3x^2?