Question about Max and Min

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elcatracho

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The equation is x^4+x^3-19x^2+32x-12. I need to find the local max and the local min. When I graph the equation on my calculator I see two humps very close to each other. The highest point is located at 1.4, 1.7 and the lowest point is located at 2,0 Now my question is are the local min and local max the highest and lowest points of the equation or are the the two spots that i have indicated. Could someone please help me with this question... ps if what I said is confusing it might be a better idea just to look at it on a calculator because it's hard to describe what I mean. Thanks so much :) Luis
 
The equation is x^4+x^3-19x^2+32x-12. I need to find the local max and the local min. When I graph the equation on my calculator I see two humps very close to each other. The highest point is located at 1.4, 1.7 and the lowest point is located at 2,0 Now my question is are the local min and local max the highest and lowest points of the equation or are the the two spots that i have indicated.
Click to expand...

Hi Luis,

To find local max and min, find the first derivative of the eqn., set it equal to zero, and solve. These are the x values of the max and min points. In general, for a fourth order eqn., you can expect three solutions – in this case, 2 mins and one max (at x = -3.8, 1.05, and 2, approx.). Plug these three x values into the original eqn to find the corresponding y values.

Hope this helps.
 
graphs

I have always been taught to find the local max and min by looking at the graphs. I do not follow what your saying but when I see the graphs I can see the humps and I have already pin pointed the Max but I am not sure which one is the min. I am looking at the graph and i'm a bit confused. All I need to the local min because i'm pretty sure I know the local max. Thanks.. :p Luis
 
You're not seeing enough of the graph. Change your window setting until you see what I mean. (You need a y-min of about -300)

for a fourth order eqn., you can expect three solutions – in this case, 2 mins and one max (at x = -3.8, 1.05, and 2, approx.). Plug these three x values into the original eqn to find the corresponding y values.
Click to expand...
 
Sorry to be a pain

I'm really sorry to bug but I thought that local max and min meant non infinity number and that there can only be one min and one max. In that case, wouldn't the lowest point on the graph be the one and only local min? I see that lowest point when I set the window to y -300 but my question is, is that point share the title of local min with the other min that appears around x=2. Thanks so much for putting up with me :) Luis
 
You are confusing yourself. The least value taken on by the function is the Global Minimum. A LOCAL minimum is just a value that is less than its neighbors. Take a trip across some rolling hills. Every time you go over a crest, that is a local maximum. IF there is one higher than ALL others, that would be the Global Maximum. Every time you hit a valley bottom, that is a local minimum. IF there is one lower than ALL others, that would be the Global Minimum.
 
ok

so in other words the local min for this problem is the point that is near the high max. It's hard for me because i've been taught that there is one max and one min and right now i'm looking at a graph that has the owest point and the highest point and thats what i've been taught. I don't know if the local min of the equation is that lowest point or the point that comes off of the highest point... like the mountain example you gave.
 
Re: graphs

elcatracho said:
I have always been taught to find the local max and min by looking at the graphs....i've been taught that there is one max and one min
Click to expand...
Then, sorry to say, you are being taught very badly.

Eliz.
 
ok

Well I may have been taught in an incorrect manner but thats why i'm asking for help because i'm confused and I don't know what I'm suppose to do. I mean the way the question is worded it says... Determine the local maxima and minima seeming to mean that they only meant one. I partially understand what people have been saying to me on here but I still can't figure out the answer. ~Luis
 
dy/dx=4x^3+3x^2-38x+32

at min dy/dx=0

0=4x^3+3x^2-38x+32

solve for x
 
Re: ok

elcatracho said:
so in other words the local min for this problem is the point that is near the high max.
Click to expand...
I'm not going to agree to that, other than to call it coincidence. Local minima and maxima do not have to be particularly close to each other. When it comes to textbook polynomials, the most interesting behavior is usually right around the Origin. This would make them close, in some sense. However, it is not hard to construct examples where this simply is not the case. There are polynomial theorems that make minima and maxima ALTERNATE, but there is no restriction on the frequency of the alternations.

Also, a global maximum (or minimum) doesn't have to exist. Examples are trivial, y = x, for example.
 
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