What are the points of inflection, if any, of #f(x) = 8 x^4−9 x^3 +9 #?

Question

What are the points of inflection, if any, of #f(x) =  8 x^4−9 x^3 +9     #?

Charles Anctil · Accepted Answer

#x=0,9/16#

The points of inflection occur when #f''(x)# switches sign.

#f(x)=8x^4-9x^3+9# #f'(x)=32x^3-27x^2# #f''(x)=96x^2-54x=6x(16x-9)#

The possible points of inflection occur when #f''(x)=0#. This is when #x=0,9/16#.

Make a sign chart that shows the potential reversals.

#color(white)(xxxxxxxxxxx)0color(white)(xxxxxxxxxxx)9/16# #larr----------------rarr# #color(white)(xxxxxx)color(red)+color(white)(xxxxxxxx)color(red)-color(white)(xxxxxxxxxxxx)color(red)+#

Since the sign changes before and after #0# and #9/16#, they are both points of inflection.

graph{[-11.71, 16.77, -0.81, 13.43]}.

Charles Arocha · Answer

undefined To find the points of inflection of $ f(x) = 8x^4 - 9x^3 + 9 $, we need to first find the second derivative, then set it equal to zero and solve for $ x $. After obtaining the values of $ x $, we can determine if they correspond to points of inflection by checking the behavior of the curvature around those points.

First derivative:
$$ f'(x) = 32x^3 - 27x^2 $$

Second derivative:
$$ f''(x) = 96x^2 - 54x $$

Setting $ f''(x) = 0 $ and solving for $ x $:
$$ 96x^2 - 54x = 0 $$
$$ 6x(16x - 9) = 0 $$

Solving for $ x $:
$$ x = 0 $$
$$ 16x - 9 = 0 $$
$$ 16x = 9 $$
$$ x = \frac{9}{16} $$

So, the points of inflection are at $ x = 0 $ and $ x = \frac{9}{16} $.