0

My class is a bit late with the material, so we didn't have a lot of time studying function derivatives, so I am having a few problems with one of the questions I was given for practising for tomorrow's math exam.

This is the question:

function: $y = x^3 - 3x^2 - 24x$

  1. Find what point resets the derivative of the function
  2. Tell the type of each points that were found in A (Minimum, Maximum, Not minimum & not maximum)
  3. Write the X and Y coordinate of when the function goes down.

So what I have tried so far:

So I know I need to derivative the function:

$y' = 3x^2 - 6x - 24$

and now I need to make 0 equal to the derivative:

$0 = 3x^2 - 6x - 24$

And then:

$\dfrac{-6 +- \sqrt{6^2 - 4 \cdot (-3) \cdot 24}} {2\cdot (-3)} => \dfrac{-6 +- 294}{2 \cdot (-3)}$

$x_1 = -48$ $x_2 = 50$

I am not sure if this is right, but what is next?

Thank you for your assistance!

user3123545
  • 123
  • You have to use the second derivative test from here. –  Apr 18 '15 at 15:40
  • If the second derivative is greater than 0 after you plug in the values of x then the point of the minima and put then value of x in the equation you get the minimum will be that value of y. It will be the absolute inverse (second derivative <0) then it is maximum –  Apr 18 '15 at 15:45
  • @Rememberme Is my answer correct? – user3123545 Apr 18 '15 at 16:05

1 Answers1

0

$y = x^3 - 3x^2 - 24x$

$y' = 3x^2 -6x - 24$

$0 = 3x^2 -6x - 24$

$\dfrac{-6 +- \sqrt{6^2 - 4 \cdot (-3) \cdot 24}} {2\cdot (-3)} => \dfrac{-6 +- 294}{2 \cdot (-3)}$

$x_1 = -48$

$x_2 = 50$

$y'' = 6x - 6$

$y_1 = -294$

$y_2 = 294$

Therefore $(-48, -294)$ is minimum point and $(50, 294)$ is a maximum point

user3123545
  • 123