critical values...

G

Guest

Guest
Hi...
My question is for the function f(x)=x^3-3x^2-x+3 find the:

a)critical values
b)for what values of x is the fcn increasing
c)for what values of x is the fcn decreasing
d)what are the coord. of the inflection points of f(x)...

So far I have found the first and second derivatives
f' = 3x^2-6x-1 and
f'' = 6x-6...

now what should I be doing??? Thanks in advance.
Chris
 
critical values are points such as the turning points(stationary) and the x and y axis intercepts.

To find the turning points take the dy/dx of your equation and let this equal 0. This is a quadratic eqn. The solutions to this give you the x value for the turning points. Place these x values (one at a time) back into the original equation and this will give you the matching y point.
 
apm,
ok if i got what you are saying correctly i set the deriv to zero and solve. I got -.1547 and 2.1547? Does that sound right?
If they are, I then plugged those in for the y portion of the points and got these (-.1547, 3.0792) & (2.1547, & -3.0792)... am I in the ballbark? If so, these are the critical values right? If they are how do I find out for what values of x is the fcn increasing and decreasing?
 
need to recheck your quadratic solutions....

f(x) =x^3 -3x^2 -x+3

f'(x) = 3x^2 -6x -1

0 = 3x^2 -6x -1

x= 1.81 or 0.18

place this back into f(x) =x^3 -3x^2 -x+3 to obtain y results.

These are the turniing points.
If you take a d^2x/dy^2 this will tell you if the turning point is a max. or min point. (positive is a min and negative is a max)

d^2x /dy^2 = 6x-6

Place the x values in this eqn and look at the result...you can then sketck the curve using the turning point and the fact it is a max. or min.


other option is to look at the dy/dx value just to the left and right of the turning point, a postive dy/dx is a rising line, negative is a falling line.

back to you.
 
having a bad calculator day today......
your quadratic solutions are correct (and the y points)

are you ok with the d^2y/dx^2 part?
 
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