How to find local maxima and minima?

Question

got my math book right here, and nothing is explained easily.
how the heck can i find the maxima and minima for, lets say,

2x^3-3x^2-36x-3 for example

THANKS IN ADVANCE!

Answer 1

Hi,

As some people aren't in calculus yet doing derivatives, you can also type the equation into your graphing calculator as
Y1 = 2x^3 -3x^2 -36x - 3. Change your window so you can see the hill and valley of the curve. Then use your minimum and maximum commands where you are looking around the hill and valley points, but be sure not to include the ends of the graph as they will get higher and lower than your relative min and max points.

You should find a minimum at (3,-84) and a relative maximum at (-2,41).

I hope that helps!! :-)


PS The first derivative explained to you also works. If you know it, the first derivative is 6x^2 - 6x - 36. If you set that equal to zero and factor it, you get 6(x - 3)(x + 2) = 0. This solves to x = 3 and x = -2. These are the x values of the relative max and min points. you'd have to plug 3 and -2 back into the original equation to get their y values.

Answer 2

Local Maxima And Minima

Answer 3

Because this is a continuous function, you can find the maxima and minima by differentiation (i.e., finding the slope).

When a curve reaches a maximum (or a minimum), its slope must be zero at that point. If the slope is stil slightly positive (not zero), then that means that the next point is higher (therefore we were not at the maximum); if the slope is slightly negative, then the point just before was higher (we were not at the maximum).

D(2x^3-3x^2-36x-3) = 6x^2 -6x -36

Set this to zero and the values for x that satisfy the requirement are critical points (could be a maximum, a minimum or an inflection point). Check each one (you'll have 2 at most).

Answer 4

you take the derivative of the function and set it equal to zero and solve for x. then you choose test points between the values and plug into the derivative to see whether it is positive or negative. where it goes from positive to negative, it is a local maxima. when it is the opposite, it is the minima. it can be either, or, or both.

ex:derivative is 6x^2-6x-36

6x^2-6x-36=0 = (6x+12)(x-3)... x=3 and x=-2



then i would choose test points 4, 1, -3 to determine their direction

f'(-3)=positive f'(-2)=zero f'(1)=negative f'(3)=zero f'(4)=positive

maxima at -2; minima at 3

Answer 5

The only way I was taught to do it was using a TI-83 graphing calculator. I can tell you how to do it using that.
1) Hit the Y= key and enter in the function
2) Hit "Trace"
3) Hit Zoom, then 0, then Trace again
4) Hit 2nd, then Trace, then 4 (for maximum)
5) On the graph, move the cursor to the left of the top of the crest of the curve you wish to find the maximum of. Hit enter.
6) Then move to the right of the crest of the same curve and hit enter.
7) Move to where you'd guess where the very top of the curve is (don't worry about getting it right on, a general area will do), and then hit enter again.
At the bottom of the screen, it'll give you the co-ordinates of the maximum.

For your problem, I got approximately (-2, 41)

The minimum is very easy to figure out from there.

I hope that this was help! :)

Answer 6

You differentiate the function and get a quadratic function which gives you the slope of the tangent of the cubic at any given point.

The turning points (ie local maxima and minima) of the cubic correspond to where slope = 0

If the slope is 0 (ie the roots or zeros of the quadratic) the tangent is flat and parallel to the x-axis and that means the curve turns and you have a local maximum or minimum.

Answer 7

This Site Might Help You.

RE:
How to find local maxima and minima?
got my math book right here, and nothing is explained easily.
how the heck can i find the maxima and minima for, lets say,

2x^3-3x^2-36x-3 for example

THANKS IN ADVANCE!

Answer 8

local maxima and minima are just the tops and bottoms of curves

Answer 9

Take the derivative and set it equal to zero. Then solve for x. This will give you the ponts where there is no slope. Meaning it is either at maximum or a minimum.
Then you take the second derivative to test if it is concave up or concave down.