How do I find concavity and points of inflection for #y = 3x^5 - 5x^3#?

Question

Evelyn Agard · Accepted Answer

At #(0, 0)# There is point of inflection
At #(1,-2)# The curve is concave upwards.
At #(-1,2)# The curve is concave downwards.

Given - #y=3x^5-5x^3# Find the first derivative - #dy/dx=15x^4-15x^2# #dy/dx =0 =>15x^4-15x^2=0 # Then - #15x^2(x^2-1)=0# #15x^2=0# #x=0# #x^2-1=0# #x^2=1# #x=+-1# #x=1# #x=-1# Find the second derivative - #(d^2x)/(dx^2)=60x^3-30x# At #x=0# #(d^2y)/(dx^2)=60(0)^3-30(0)=0# The value of the function - #y=3(0)^5-5(0)^3=0# At #(0, 0)# #dy/dx=0; (d^2y)/(dx^2)=0# Hence there is a point of inflection at #(0, 0)# At #x=1# #(d^2y)/(dx^2)=60(1)^3-30(1)=60-30=30>0# The value of the function - #y=3(1)^5-5(1)^3=03-5=-2# At #(1, -2)# #dy/dx=0; (d^2y)/(dx^2)>0# Hence there is a minimum at #(1, -2)# The curve is concave upwards At #x=-1# #(d^2y)/(dx^2)=60(-1)^3-30(-1)=-60+30=-30<0# The value of the function - #y=3(-1)^5-5(-1)^3=-3+5=2# At #(1, 2)# #dy/dx=0; (d^2y)/(dx^2)<0# Hence there is a Maximum at #(1, 2)# The curve is concave downwards.

Paisley Johnson · Answer

undefined To find the concavity and points of inflection for $ y = 3x^5 - 5x^3 $, follow these steps:

1. Find the second derivative of the function $ y $.
2. Set the second derivative equal to zero and solve for $ x $ to find any possible points of inflection.
3. Determine the sign of the second derivative on intervals between these points to identify the concavity of the function.

Let's proceed with these steps:

1. First derivative: $ y' = 15x^4 - 15x^2 $
2. Second derivative: $ y'' = 60x^3 - 30x $
3. Set $ y'' $ equal to zero and solve for $ x $:
   $ 60x^3 - 30x = 0 $
   Factor out $ 30x $: $ 30x(x^2 - 1) = 0 $
   Solve for $ x $:
   $ x = 0 $ (critical point), $ x = -1 $, $ x = 1 $
4. Determine the sign of $ y'' $ in each interval:
   - Interval $ (-\infty, -1) $: Pick $ x = -2 $ (test point), $ y''(-2) = -120 < 0 $ so concave down.
   - Interval $ (-1, 0) $: Pick $ x = -0.5 $ (test point), $ y''(-0.5) = 15 > 0 $ so concave up.
   - Interval $ (0, 1) $: Pick $ x = 0.5 $ (test point), $ y''(0.5) = 15 > 0 $ so concave up.
   - Interval $ (1, \infty) $: Pick $ x = 2 $ (test point), $ y''(2) = 120 > 0 $ so concave up.

Points of inflection: $ x = -1 $ and $ x = 1 $
Concave up: $ (-1, 0) $ and $ (0, 1) $
Concave down: $ (-\infty, -1) $ and $ (1, \infty) $