Finding Max and Mins

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Hckyplayer8

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Find all three critical numbers of f(x)=(x-1)2(x-2)3

Fermat's Theorm makes sense. Visually the slope of the tangent line is 0...fully horizontal...at the max and min of the function. Looking for a critical number means I'm looking for where the slope is 0 or does not exist.

In my teachers example, he had x on an interval which this problem doesn't give...at least to my untrained eye. Outside of randomly plugging in numbers, how does one solve this?
 
Hckyplayer8 said:
Find all three critical numbers of f(x)=(x-1)2(x-2)3

Fermat's Theorm makes sense. Visually the slope of the tangent line is 0...fully horizontal...at the max and min of the function. Looking for a critical number means I'm looking for where the slope is 0 or does not exist.

In my teachers example, he had x on an interval which this problem doesn't give...at least to my untrained eye. Outside of randomly plugging in numbers, how does one solve this?
Click to expand...
How did the teacher find points of 0 slope tangents?
 
Hckyplayer8 said:
Find all three critical numbers of f(x)=(x-1)2(x-2)3

Fermat's Theorm makes sense. Visually the slope of the tangent line is 0...fully horizontal...at the max and min of the function. Looking for a critical number means I'm looking for where the slope is 0 or does not exist.

In my teachers example, he had x on an interval which this problem doesn't give...at least to my untrained eye. Outside of randomly plugging in numbers, how does one solve this?
Click to expand...
Can you calculate the first derivative of the given function?
 
All you need to do is to take the derivative, set it equal to zero, and solve that equation. Don't take derivatives of derivatives!

The question is, what is the derivative of f(x)=(x-1)2(x-2)3 ? Please show what you can do with that, and we can continue from there.

Note that the domain of f is all real numbers.
 
Dr.Peterson said:
All you need to do is to take the derivative, set it equal to zero, and solve that equation. Don't take derivatives of derivatives!

The question is, what is the derivative of f(x)=(x-1)2(x-2)3 ? Please show what you can do with that, and we can continue from there.

Note that the domain of f is all real numbers.
Click to expand...

Okay. First is the product rule which is d1 * f2 + f1 *d2.

So 2(x-1) * (x-2)3 + (x-1)2 * 3(x-2)2

I feel like the chain rule would also come to play because is this not a composition as well?
 
It could have, but the derivatives of the "inside functions" x-1 and x-2 are just 1.

Now continue: write and solve the equation 2(x-1) * (x-2)3 + (x-1)2 * 3(x-2)2 = 0. You'll want to do some factoring.
 
Dr.Peterson said:
It could have, but the derivatives of the "inside functions" x-1 and x-2 are just 1.

Now continue: write and solve the equation 2(x-1) * (x-2)3 + (x-1)2 * 3(x-2)2 = 0. You'll want to do some factoring.
Click to expand...

After factoring out the above I'm left with (2x-2) * (x3-3x2 * 2 +3x * 22-23) * (x2-x-x+1) * 3x2-6x-6x+12) =0

Which then simplifies to (2x-2) * (x3-6x2+12x-8) * (3x2-12x+12)=0
 
Can you differentiate the function?
Hckyplayer8 said:
After factoring out the above I'm left with (2x-2) * (x3-3x2 * 2 +3x * 22-23) * (x2-x-x+1) * 3x2-6x-6x+12) =0

Which then simplifies to (2x-2) * (x3-6x2+12x-8) * (3x2-12x+12)=0
Click to expand...
You've expanded the expression rather than factorising it.

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

Notice that each TERM (the 2 terms are the things before and after the + sign in the middle) has a factor of (x-1) and (x-2)2.
So:

(x-1)(x-2)2 [2(x-2)+3(x-1)] = 0

Tidy up inside the square bracket. Can you finish it now?
 
Harry_the_cat said:
Can you differentiate the function?

You've expanded the expression rather than factorising it.

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

Notice that each TERM (the 2 terms are the things before and after the + sign in the middle) has a factor of (x-1) and (x-2)2.
So:

(x-1)(x-2)2 [2(x-2)+3(x-1)] = 0

Tidy up inside the square bracket. Can you finish it now?
Click to expand...

Crap. I don't know how people remember all this stuff...
 
Perhaps its because my brain is shot after 14 weeks of this subject. Perhaps its because its 1am. I've sat and looked at the factoring portion and I don't see it. Can somebody show me the breakdown of that?
 
Now you need to find which one is max or min?
Be careful though, if derivative at a point is zero that doesn’t necessarily mean you have max or min there
For example: y=x^3
First derivative =0 —-> 3x^2=0 —> x=0
Even though the slope (derivative) is 0 at x=0 that doesn’t mean we got a max or min
You either take the second derivative or give x values smaller and bigger and check the sign of derivative (slope)
Like here in x^3.... the derivative or slope will be 3x^2.... if i give -0.1 for x, the slope will be positive if i give 0.1 again it will be positive
The slope didn’t change and we know for min max point we should have change in slope
 
Hckyplayer8 said:
Perhaps its because my brain is shot after 14 weeks of this subject. Perhaps its because its 1am. I've sat and looked at the factoring portion and I don't see it. Can somebody show me the breakdown of that?
Click to expand...
Can you see that factorising 2ab3 + 3a2b2 = ab2(2b+3a) ?

So if a = (x-1) and b = (x-2), that's exactly what I have done in my previous post.
 
Harry_the_cat said:
Can you see that factorising 2ab3 + 3a2b2 = ab2(2b+3a) ?

So if a = (x-1) and b = (x-2), that's exactly what I have done in my previous post.
Click to expand...

Now I got it. Thanks.

The rest of the problem ends up being fairly simple...I hope.

x cords that equal zero are 1, 2 and 7/5 and that should be it.

Thank you all for posting.
 
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