# Thread: Critical points of a function

1. ## 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?

2. Hello,

why is it 18-5 ? Is there any typo here ? Because it's strange it hasn't been simplified into 13
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.

3. 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
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.

4. hey there
you'll have

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

5. 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.

6. 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''...

7. 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.