minimization - why do you substitute 0 for f ' (x)

[Kristy]

minimization - why do you substitute 0 for f ' (x)

Number 2.
Directions:
Find two numbers whose difference is 100 and whose product is a minimum.

My work:
Let x be the first number.
Let x + 100 be the second number.

Minimize x * (x+100)
f (x) = \(\displaystyle x^{2} + 100x\)

f ’(x) = 2x + 100

Substitute 0 for f ‘ (I’m not exactly sure why)
0 = 2x + 100
-100 = 2x
-50 = x

So the numbers are -50 and 50.

Does that seem right at all? I can't figure out what I'm doing wrong.

 

[galactus]

The reason for that, Kristy, is max and mins occur where the slope is 0. That is, the tangent line is horizontal.

Suppose you have f(x)=x^3+2x^2+1.

Then f'(x)=3x^2+2x. If you set this to 0 and solve for x, that's the places where the slope is 0. Those are the max or min values. High and low points, if you will. See what I mean?.

In this one you'll find there are two places where this occurs.

 

[Kristy]

The reason for that, Kristy, is max and mins occur where the slope is 0. That is, the tangent line is horizontal.

Suppose you have f(x)=x^3+2x^2+1.

Then f'(x)=3x^2+2x. If you set this to 0 and solve for x, that's the places where the slope is 0. Those are the max or min values. High and low points, if you will. See what I mean?.

In this one you'll find there are two places where this occurs.

Ok, I do understand the max and min thing now. That makes sense because the derivative is the slope of the tangent line. I just try to picture a max and it seems like it would have a flat tangent line (but then I was also picturing something like the absolute value function, does it have a min at 0? I know its graph looks like a V with the lowest poitn at 0, but would there even be a tangent line there?

On your problem.
0 = 3x^2 + 2x
0 = x (3x + 2)
x = 0 or 3x+2 = 0
3x = -2
x = -2/3

So I see where there are two answers.

Thanks!

 

[galactus]

Absolute value has that 'peak'. It's not differentiable there.

Here's the analytic proof (though you can see it geometrically) if you want to see it. If you don't, here it is anyway

We will show that f(x)=|x| is not differentiable at x=0.

\(\displaystyle \L\\f'(0)=\lim_{h\to\0}\frac{f(0+h)-f(0)}{h}\\=\lim_{h\to\0}\frac{f(h)-f(0)}{h}\\=\lim_{h\to\0}\frac{|h|-|0|}{h}\\=\lim_{h\to\0}\frac{|h|}{h}\)

But, \(\displaystyle \L\\\frac{|h|}{h}=1, \;\ for \;\ h>0\)

and \(\displaystyle \L\\\frac{|h|}{h}=-1, \;\ for \;\ h<0\)

So that:

\(\displaystyle \L\\\lim_{h\to\0^{-}}\frac{|h|}{h}=-1\)

and \(\displaystyle \L\\\lim_{h\to\0^{+}}\frac{|h|}{h}=1\)

Therefore, \(\displaystyle \L\\f'(0)=\lim_{h\to\0}\frac{|h|}{h}\) does not exist because the one-sided limits are not the same. So, \(\displaystyle f(x)=|x|\) is not differentiable at x=0.

There, wasn't that fun?.

Next time you go to a party, you can impress your friends by showing the non-differentiablity of |x| at x=0. :wink:

 

[Kristy]

Absolute value has that 'peak'. It's not differentiable there.

Here's the analytic proof (though you can see it geometrically) if you want to see it. If you don't, here it is anyway

We will show that f(x)=|x| is not differentiable at x=0.There, wasn't that fun?.

Next time you go to a party, you can impress your friends by showing the non-differentiablity of |x| at x=0. :wink:

Cool. Good to know that it isn't differntiable at zero (and I think I even understand it! I will keep that stored away for some day when the professor asks us to write "something that we've learned that wasn't on the test" - I seriously had one that did that in a non math class. or the next time I need to totally confuse someone, just kidding.

So that's good to know that some functions are NOT differentiable at their min or max. What do you do in that case? Give up? Draw a graph? Become a math major?

 

[galactus]

Become a math major. There are lot's of interesting things with regard to continuity and differentiablility.

For instance, |x| is continuous, but not differentiable at x=0.

A function can be continuous but not differentiable. But, if it is differentiable, then it has to be continuous.

|x| is a prime example.

Also, a continuous function can have a derivative that's not continuous.

Again, |x|.

If x>0, then f(x)=|x|=x, f'(x)=1. If x<0, then f(x)=|x| = -x, so f'(x) = -1.

\(\displaystyle \L\\f'(x)=\frac{d}{dx}[|x|]=\begin{Bmatrix}1, \;\ x>0\\-1, \;\ x<0\end{Bmatrix}\)

So, f' is not continuous.

You can see by the graph.

 

[Kristy]

Cool! So if it was one of those Venn diagrams,
kinda like dog vs animal
continuity would be an outer circle, (like the category animal)
and differentiability would be a small circle inside the continuity circle? (like the specifc type dog)

As in if its differeential it automatically is continuous, (dog is always an animal)
but a continuous function doesn't have to be differentiable (an animal doesn't have to be a dog?

(I thought that was a cool way to remember it because of the D in Dog and Differentiable.)

Nope, just calc 1. Aparently you don't need to know multivariable calculus to be a vet.