How do you use the second derivative test to find min/max/pt of inflection of #y= x^5-5x#?

Question

Emily Ammerman · Accepted Answer

Maximum at #x=-1 and y=4#
Minimum at #x=1 and y=-4#
Inflection point at #x=0 and y=0#.

Start by taking the first derivative:
#y'=5x^4-5#
set it equal to zero:
#5x^4-5=0#
#x=+-1# corresponding to #y=+-4#; these will be the #x#-coordinates of maxima or minima!

Now we find the second derivative:
#y''=20x^3#
we analyze the sign of the second derivative by setting it bigger than zero:
#20x^3>0#
so this is true when #x>0#
we have an inflection point at #x=0# where your function changes concavity;
Graphically:

Elijah Abram · Answer

undefined To use the second derivative test to find the minimum, maximum, or point of inflection of $ y = x^5 - 5x $, follow these steps:

1. Find the first derivative of the function $ y = x^5 - 5x $ to determine critical points.
   $$ y' = 5x^4 - 5 $$
2. Set the first derivative equal to zero and solve for $ x $ to find the critical points.
   $$ 5x^4 - 5 = 0 $$
   $$ x^4 - 1 = 0 $$
   $$ (x^2 - 1)(x^2 + 1) = 0 $$
   $$ x = \pm 1 $$
3. Determine the sign of the second derivative at each critical point to classify them.
   $$ y'' = 20x^3 $$
   At $ x = -1 $:
   $$ y''(-1) = -20 $$
   This indicates a local maximum.
   At $ x = 1 $:
   $$ y''(1) = 20 $$
   This indicates a local minimum.
4. Check the concavity of the function to identify points of inflection.
   Since the function changes concavity at $ x = 0 $, it is a point of inflection.

So, for $ y = x^5 - 5x $:
- $ (1, -4) $ is a local minimum.
- $ (-1, 4) $ is a local maximum.
- $ (0, 0) $ is a point of inflection.