critical values?

Question

how many critical values and how many points of inflection does y=x^3-3x^5 have?

Answer 1

To solve for how many critical values it has, you must first find the derivative and then make it 0.

y' = 3x^2 - 15x^4
0 = 3x^2 - 15x^4

0 = 3x^2 (1 - 5x^2)

3x^2 = 0
1 - 5x^2 = 0

x^2 = 0
5x^2 = 1

x = 0
x^2 = 1/5

x = 0
x = +/- 1/sqrt(5)

Critical values: 0, 1/sqrt(5), -1/sqrt(5)
So it has 3 critical values.

To get the points of inflection, you have to take the second derivative and make it 0.

y'' = 6x - 60x^3
0 = 6x - 60x^3
0 = 6x (1 - 10x^2)

6x = 0
1 - 10x^2 = 0

x = 0
10x^2 = 1

x = 0
x^2 = 1/10

x = 0
x = +/- 1/sqrt(10)
Points of inflection: 0, 1/sqrt(10),-1/sqrt(10)
Therefore, there are 3 point of inflection.