Local minima and maxima, critical/inflection points

[EETman]

Let f(x)=x⁴ - 4x<sup>3

When differentiated i get f'(x)=4x^3-12x^2
which can factor to the form 4x^2(x-3)

so based off of this assumption critical points lie at 3 and 0, however when i graph it on my calculator the critical points seem to lie around +/- 2.44 as well as 0.
any help would be appreciated.

EDIT: its obvious that the answer is +/- SQRT(6) , i just dont see how to arrive at that answer from where i am.</sup>

 

[Romsek]

Let f(x)=x⁴ - 4x<sup>3

When differentiated i get f'(x)=4x^3-12x^2
which can factor to the form 4x^2(x-3)

so based off of this assumption critical points lie at 3 and 0, however when i graph it on my calculator the critical points seem to lie around +/- 2.44 as well as 0.
any help would be appreciated.

EDIT: its obvious that the answer is +/- SQRT(6) , i just dont see how to arrive at that answer from where i am.</sup>


View attachment 3448

You must have entered it into your calculator incorrectly.

The plot shows both f(x) = x^4 - 4 x^3, and it's derivative f'(x) = 4 x^3 - 12 x^2.

It's pretty clear from the plot that the zeros of the derivative are at 0 and 3 which also correspond with a local maxima and a local minima of f

 

[EETman]

Let f(x)=x⁴ - 4x<sup>3

When differentiated i get f'(x)=4x^3-12x^2
which can factor to the form 4x^2(x-3)

so based off of this assumption critical points lie at 3 and 0, however when i graph it on my calculator the critical points seem to lie around +/- 2.44 as well as 0.
any help would be appreciated.

EDIT: its obvious that the answer is +/- SQRT(6) , i just dont see how to arrive at that answer from where i am.</sup>


View attachment 3448

You must have entered it into your calculator incorrectly.

The plot shows both f(x) = x^4 - 4 x^3, and it's derivative f'(x) = 4 x^3 - 12 x^2.

It's pretty clear from the plot that the zeros of the derivative are at 0 and 3 which also correspond with a local maxima and a local minima of f

It was outside the window, thanks!

 

[JeffM]

Let f(x)=x⁴ - 4x<sup>3

When differentiated i get f'(x)=4x^3-12x^2
which can factor to the form 4x^2(x-3)

so based off of this assumption critical points lie at 3 and 0, however when i graph it on my calculator the critical points seem to lie around +/- 2.44 as well as 0.
any help would be appreciated.

EDIT: its obvious that the answer is +/- SQRT(6) , i just dont see how to arrive at that answer from where i am.</sup>

\(\displaystyle f(x) = x^4 - 4x^3 \implies f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \implies f'(x) \ne 0\ unless\ x = 0\ or\ x = 3.\)

So you got that right. You do not tell us what the problem asked you to do so it is hard to tell where you are having difficulty. In the future please give us the problem exactly.

Clearly, your graph is not of f(x).

Now if you are supposed to find the values of x where f(x) has finite extrema (maxima and minima), then MAYBE such values of x are either 0 or 3. We determine that by looking at higher order derivatives.

\(\displaystyle f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \implies f''(x) = 12x^2 - 24x = 12x(x - 2) \implies f''(0) = 0\ and\ f''(3) = + 36.\)

Because the second derivative at x = 3 is positive, x = 3 is at least a local minimum. (In fact, it is a global minimum although that can not be determined simply by looking at critical points.) But f''(0) = 0, which means that 0 may be an inflection point. If x < 0, f'(x) < 0. And if
0 < x < 3, f'(x) < 0. So 0 is an inflection point.

 

[lookagain]

\(\displaystyle f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \implies f''(x) = 12x^2 - 24x = 12x(x - 2) \implies f''(0) = 0\ and\ f''(3) = + 36.\)


But f''(0) = 0, which means that 0 may be an inflection point. If x < 0, f'(x) < 0.

And if 0 < x < 3, f'(x) < 0. So 0 is an inflection point.

From setting f''(x) = 0, x = 0 and x = 2 are second order critical numbers. They are x-coordinates of

potential/possible inflection points.


A student should be using test numbers on either side of these critical numbers and substitute those into the

second derivative to see if concavity changes from one side of a critical number to the other side of it.


For example, pick test numbers -1, 1, and 3, and substitute them into the second derivative.


On either side of x = 0, the sign of the second derivative changes, and on either side of x = 2, the sign of the

second derivative also changes.


So, inflection points occur at x = 0 and at x = 2. To complete the coordinates of the inflection points,

substitute the respective x-values into f(x) to get the corresponding y-values.