jazzycheri jazzycheri
  • 04-01-2017
  • Mathematics
contestada

How do I find the critical points?

[tex]( x^{2} - 1)^3 [/tex]

Intervals: [ -1, 2]

Respuesta :

Hilltop1108 Hilltop1108
  • 04-01-2017
Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:

[tex] \frac{d}{dx}(x^{2} - 1)^{3} [/tex]

So we apply chain rule:

= [tex]3(x^{2} - 1)^{2} * 2x[/tex]

Set our first derivative to zero and solve for x:

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

So we can see that (by plugging in) 0, -1 and 1 makes our solution true

So our critical value is x = 0, x = -1, x = 1
Answer Link

Otras preguntas

nucleotides are the basic unit of which macromolecules
Which of these statements is true for f(x)=4*(7)^x?
d+(−5.004)=2.826 I am so confused. What do I do?
how do u wright 3.3 in rational numblersl
Who discovered skin cells
Please help! 1. Complète la phrase avec quel, quelle, quels, quelles, lequel, laquelle, lesquels ou lesquelles. _________ voiture va plus vite? 2. Complète la p
Please help with the question
the last choice is 75 plz help
Hello can somone help me ? Copper has an atomic number of 29. That means that there are 29 protons in the nucleus of each copper atom. That's what the atomic nu
how would isopropanol behave when poured out of its container onto a table