How do you find inflection points #f(x) = x^4 - 2x^2 + 4#?

Question

Isabella Ackerman · Accepted Answer

They occur at #x=\pm\frac{1}{\sqrt{3}}#.  On the graph of #f#, they are at the points #(x,y)=(\pm\frac{1}{\sqrt{3}},f(\pm\frac{1}{\sqrt{3}}))=(\pm\frac{1}{\sqrt{3}},31/9)#. The first derivative is #f'(x)=4x^{3}-4x# and the second derivative is #f''(x)=12x^{2}-4=4(3x^{2}-1)#.  Setting #f''(x)=0# gives possible inflection points at #x=\pm\frac{1}{\sqrt{3}}#.  That these are actual inflection points follows from the fact that #f''(x)# changes sign at these values of #x#.  The corresponding #y#-coordinate of both of the inflection point is #f(\pm\frac{1}{\sqrt{3}})=31/9#.

Ezekiel Alexander · Answer

undefined To find the inflection points of $ f(x) = x^4 - 2x^2 + 4 $, follow these steps:

1. Compute the second derivative of the function $ f(x) $, denoted as $ f''(x) $.
2. Set $ f''(x) = 0 $ and solve for $ x $.
3. Find the corresponding $ y $-coordinates for the $ x $-values obtained in step 2.
4. The points obtained in step 3 are the inflection points of the function.

First, compute the second derivative:
$$ f'(x) = 4x^3 - 4x $$
$$ f''(x) = 12x^2 - 4 $$

Now, set $ f''(x) = 0 $ and solve for $ x $:
$$ 12x^2 - 4 = 0 $$
$$ 12x^2 = 4 $$
$$ x^2 = \frac{4}{12} $$
$$ x^2 = \frac{1}{3} $$
$$ x = \pm \sqrt{\frac{1}{3}} $$

Now find the corresponding $ y $-coordinates:
$$ f\left(\sqrt{\frac{1}{3}}\right) = \left(\sqrt{\frac{1}{3}}\right)^4 - 2\left(\sqrt{\frac{1}{3}}\right)^2 + 4 $$
$$ f\left(\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{2}{3} + 4 $$
$$ f\left(\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{6}{9} + 4 $$
$$ f\left(\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{6}{9} + \frac{36}{9} $$
$$ f\left(\sqrt{\frac{1}{3}}\right) = \frac{31}{9} $$

$$ f\left(-\sqrt{\frac{1}{3}}\right) = \left(-\sqrt{\frac{1}{3}}\right)^4 - 2\left(-\sqrt{\frac{1}{3}}\right)^2 + 4 $$
$$ f\left(-\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{2}{3} + 4 $$
$$ f\left(-\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{6}{9} + 4 $$
$$ f\left(-\sqrt{\frac{1}{3}}\right) = \frac{1}{9} - \frac{6}{9} + \frac{36}{9} $$
$$ f\left(-\sqrt{\frac{1}{3}}\right) = \frac{31}{9} $$

So, the inflection points are $ \left(\sqrt{\frac{1}{3}}, \frac{31}{9}\right) $ and $ \left(-\sqrt{\frac{1}{3}}, \frac{31}{9}\right) $.