First time one-need help

[shy5577]

Here's the problem I'm trying to work out. Pls help

Y=X^3-6X^2-15X+4

Find critical pts
intervals where y is increasing and decreasing
max and min values of y
any pts of inflection

Here's what I have

(-1,2) critical pts
(-1,12) rel max
(5,-96) rel min
(2,-42) pts of infle

 

[stapel]

I'm sorry, but I don't follow the "what I have" part of your post. For instance, aren't max/min points also critical points? So how are you getting max/min points that aren't critical points? Or am I misunderstanding what you've posted...?

To find the critical points, find the derivative and set the result equal to zero. In your case, since you're starting out with a cubic, the derivative, naturally, will be a quadratic. So you can use the Quadratic Formula, if you like, to find the two critical points.

Have you learned about the Second Derivative Test yet? Or are you using other methods to find where the function is increasing or decreasing?

The posted exercise asks for the "max and min values of y". Does this mean "global" max and min, or local max and mix? Or both? If the former, over what interval?

I agree with the only point of inflection being at x = 2, y = -42. Regarding the rest of the exercise, please pardon my confusion and reply with clarification. Thank you.

Eliz.

 

[Daniel_Feldman]

Ok....we are given

Y=X^3-6X^2-15X+4


First order of business is to derive it:

y'=3x^2-12x-15
=3(x^2-4x-5)
=3(x-5)(x+1)

Now, critical numbers are the values of x where the derivative is equal to 0 (or where the derivative does not exist, but that does not concern us here, as the function is differentiable everywhere).

So, 3(x-5)(x+1)=0

x=5
x=-1

When x=5, y is equal to -96.
When x=-1, y is equal to 12.

So, the critical points are (5,-96) and (-1,12).


Now, we need to do what's called a factor line analysis; That is, wer have to plot our two x coordinated of the critical numbers (5 and -1) on a number line and find where the first derivative is positive and where it is negative (this is known as the first derivative test).

For x<-1, y'>0. For -1<x<5, y'<0. For x>5, y'>0.

The function is increasing on the open interval(s) where the derivative is positive, and decreasing on the open interval(s) where the derivative is negative.

Therefore, this function is increasing on (-infinity,-1) U (5,infinity) and decreasing on (-1,5).

A function has a relative min at the x values where y' changes from negative to positive, and a relative max where y' changes from positive to negative. From our work above, where/what would those be?

A point of inflection occurs where the function changes concavity; That is, where the second derivative changes sign. So we merely find the second derivative and do a factor line analysis on it. Try doing it from our work above.

 

[shy5577]

Thanks Daniel

So my critical pts are (5,-96) (-1,12)
Rel min (5,-96)
rel max (-1,12)
pts of inflection (2,-42)

So, my second derivative is
6x-12=0
x=2

This is how I set up for both derivatives

x<-1 -1<x<2 2<x<5 x>5
f1(x)= increase decrease decrease increase
f2(x)= decrease decrease increase increase
f(x)= U U

[list][/list]

 

[Daniel_Feldman]

So (2,-42) is the only point of inflection.