What are the points of inflection of #f(x)=3x^5 - 5x^3 ##?

1 Answer
Jan 4, 2016

They are
# (-sqrt2/2, sqrt2/4) , (0, 0) , (1, -sqrt2/4)#

Explanation:

Possible point of inflection is determine by setting the derivative equal to zero. Possible point on inflection happen when there is changes in concavity (refer to the graph below)

#f(x) = 3x^5 -5x^3#

#f'(x) = 15x^4 -15x^2#

#f''(x) = 60x^3 -30x#

Set the second derivative equal to zero

#60x^3 - 30x = 0#

#30x(2x^2 -1) = 0#

#30x= 0 => x = 0#

#x^2-1 = 0=> 2x^2= 1#

#x^2 =1/2>x = +-sqrt(1/2) => +- sqrt(2)/2#

Coordinates point for P.O.I (Point of inflection)

#f(-sqrt2/2) = 3(-sqrt2/2)^5 - 5(-sqrt2/2) #

#(3)(-sqrt2/8) -5(-sqrt2/8)=-3sqrt2/8+5sqrt2/8 #

#2sqrt2/8=>sqrt2/4==> (-sqrt2/2,sqrt2/4)#

#f(0) = 0-0 =0 => (0, 0)#

#f(sqrt2/2) = 3(sqrt2/2)^5 - 5(sqrt2/2) #

#(3)(sqrt2/8) -5(sqrt2/8)=3sqrt2/8-5sqrt2/8 #

#-2sqrt2/8=>-sqrt2==> (-sqrt2/2,-sqrt2/4)#

graph{3x^5-5x^3 [-10, 10, -5, 5]}