# Critical points of a function

• Jun 9th 2009, 11:30 AM
sinewave85
Critical points of a function
Using the first and second order derivatives to find the critical points of this function has left me very confused -- they don't mesh at all with the critical points on the graph of the function (as displayed by a graphing calculator).

$f(x) = 3x^{4} - 2x^{3} - 12u^{2} + 18x -5$

$f^{\prime}(x) = 12x^{3} - 6x^{2} - 24x + 18$

$f^{\prime}(x) = 6(2x-2)(x-1)(x+1.5)$

$f^{\prime \prime}(x) = 36x^{2} - 12x -24$

$f^{\prime \prime}(x) = 12(3x + 2)(x -1)$

So, that give potential critical points of x= 1, and -3/2 for relative extremas and x=-2/3 for a point of inflection.

The calculator displays critical points (min, max, min, respectively) of x= 3, 0.6861393, and -2.186138.

Why the discrepancy?
• Jun 9th 2009, 11:35 AM
Moo
Hello,

why is it 18-5 ? Is there any typo here ? Because it's strange it hasn't been simplified into 13 (Surprised)
And hence it seems not logical that the derivative has 18 in it...

Also, what is u ? If it doesn't depend on x, then its derivative is 0 and you shouldn't have any u left in the derivative of f.
• Jun 9th 2009, 11:39 AM
sinewave85
Quote:

Originally Posted by Moo
Hello,

why is it 18-5 ? Is there any typo here ? Because it's strange it hasn't been simplified into 13 (Surprised)
And hence it seems not logical that the derivative has 18 in it...

Also, what is u ? If it doesn't depend on x, then its derivative is 0 and you shouldn't have any u left in the derivative of f.

Ok, I must not have my thinking cap on today. The 18 is missing its variable -- supposed to be 18x. And the u's are just a typo, too. The book uses u for the variable, and I was changing it to x as I typed (just because that is the standard variable) and missed those. I fixed the op.
• Jun 9th 2009, 12:04 PM
javax
hey there
you'll have

$y' = 6(2x^3-x^2-4x+3) = 6(2x+3)(x-1)^2$
• Jun 9th 2009, 12:10 PM
sinewave85
Quote:

Originally Posted by javax
hey there
you'll have

$y' = 6(2x^3-x^2-4x+3) = 6(2x+3)(x-1)^2$

Hi, javax. The thing is, that gives the same solutions as the way I have it written out, right:

$y' = 6(2x+3)(x-1)^2$

$0=(2x+3)(x-1)(x-1)$

$x=\frac{-3}{2}, 1$

So we are still facing the same question.
• Jun 9th 2009, 12:23 PM
javax
If we're talking about $y = 3x^4-2x^3-12x^2+18x-5$
then it's derivative is
$y' = 6(2x+3)(x-1)^2$ which is negative in the interval $(-\infty, -\frac{3}{2})$thus the function decreasing in this interval and it will be positive in the interval $(-\frac{3}{2}, \infty)$ which means the function is increasing in this interval. I'm sure this is true for this function.
The same procedure with y''...
• Jun 9th 2009, 12:30 PM
sinewave85
Quote:

Originally Posted by javax
If we're talking about $y = 3x^4-2x^3-12x^2+18x-5$
then it's derivative is
$y' = 6(2x+3)(x-1)^2$ which is negative in the interval $(-\infty, -\frac{3}{2})$thus the function decreasing in this interval and it will be positive in the interval $(-\frac{3}{2}, \infty)$ which means the function is increasing in this interval. I'm sure this is true for this function.
The same procedure with y''...

Ok, thanks. I think something was set incorrectly in my calculator and it was creating an inaccurate graph. Thanks for the help.