How do you find all points of inflection given #y=-x^5+2x^3+4#?

Question

Charles Arocha · Accepted Answer

Inflection points at #x = 0# and #x = +-sqrt(5/3)#

Find the second derivative:

#y' = -5x^4 + 6x^2#

#y'' = -20x^3 + 12x#

This will have points of inflection when #y'' =0 #.

#0 = -20x^3 + 12x#

#0 = -4x(5x^2 - 3)#

This means that #x = 0# or #x = +-sqrt(5/3)#.

Hopefully this helps!

Luke Alvarez · Answer

undefined To find all points of inflection for the function $ y = -x^5 + 2x^3 + 4 $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $ to find potential points of inflection.
3. Test each potential point of inflection by checking the sign of the second derivative around that point.
4. Points where the second derivative changes sign are the points of inflection.

Let's go through these steps:

1. Find the second derivative of $ y $:
   $ y' = -5x^4 + 6x^2 $
   $ y'' = -20x^3 + 12x $

2. Set the second derivative equal to zero and solve for $ x $:
   $ -20x^3 + 12x = 0 $
   $ 4x(-5x^2 + 3) = 0 $
   $ x = 0 $ or $ x = \pm \sqrt{\frac{3}{5}} $

3. Test each potential point of inflection:
   - For $ x = 0 $, evaluate $ y''(x) $ for values slightly less and slightly more than 0.
   - For $ x = \pm \sqrt{\frac{3}{5}} $, evaluate $ y''(x) $ for values slightly less and slightly more than $ \pm \sqrt{\frac{3}{5}} $.

4. Determine the sign of $ y''(x) $ around each potential point of inflection. Points where the sign changes indicate points of inflection.

Therefore, the points of inflection for the function $ y = -x^5 + 2x^3 + 4 $ occur at $ x = 0 $ and $ x = \pm \sqrt{\frac{3}{5}} $.