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

1 Answer
Nov 10, 2016

There is just one point of inflection at the origin

Explanation:

# y = x^5 - 2x^3 #

Differentiating wrt #x# we have'

# dy/dx = 5x^4 - 6x^2 #

At a critical point , # dy/dx = 0 #
# dy/dx = 0 => 5x^4 - 6x^2 = 0 #
# :. x^2(5x^2 - 6) = 0 #
So, Either # x^2 = 0 => x= 0#, or # 5x^2 - 6 = 0 => x =+- sqrt(6/5) #

So critical points occurs when #x=0, x =+- sqrt(6/5)#

Next we can find the nature of th critical points by looking at the second derivative:

# dy/dx = 5x^4 - 6x^2 #
# :. (d^2y)/(dx^2) = 20x^3 - 12x #
# :. (d^2y)/(dx^2) = 4x(5x^2 - 3) #

When # { (x=-sqrt(6/5), => (d^2y)/(dx^2)<0,=>,"maximum"), (x=0, => (d^2y)/(dx^2)=0,=>,"inflection"), (x =sqrt(6/5), => (d^2y)/(dx^2)>0,=>,"minimum") :} #

So therei is one point of inflection when #x=0 => y=0#

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