Prove that x^2 is always positive.
I understand why, but I don't know how to write out.

- anonymous

Prove that x^2 is always positive.
I understand why, but I don't know how to write out.

- katieb

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- Mertsj

Try an indirect proof....assume that it is negative and show that it leads to a contradiction.

- nenadmatematika

for every x in R y=x^2 stays equal to zero or bigger than zero

- myininaya

what do you mean you don't know how to write x^2>0?

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## More answers

- myininaya

and yes because one counterexample for all complex would be if x=i
i^2=-1<0

- nenadmatematika

correct....just look out because it could be equal to zero when x=0

- myininaya

yep yep

- myininaya

so the best thing to say is x^2>=0 for all reals

- Zarkon

show thats its minumm value is 0

- myininaya

which nenad already said

- Zarkon

I butchered the crap out of that word....minimum ;)

- myininaya

zarkon you don't have to know how to spell

- myininaya

you just have to get it a little right

- Zarkon

good...because I can't

- anonymous

Here is the problem:
f'(x) = (x-a)(x-b)^2
0 < a < b
"For which values for a and b is f increasing?"
And I can't motivate why (x-b)^2 is always positive.
Am I over thinking this?

- myininaya

f'=2x=0 => x=0
which means x=0 is a critical number
f''=2 which means it is concave up

- myininaya

so a and b are critical numbers
\[f''=(x-b)^2+(x-a) \cdot 2(x-b)=(x-b)[x-b+2(x-a))]=(x-b)[3x-b-2a]\]
\[f''(a)=(a-b)[3a-b-2a]=(a-b)[a-b]=(a-b)^2=(b-a)^2>0\]
\[f''(b)=0 \text{ inconclusive }\]

- myininaya

b>a
so b is not a ever
so (b-a)^2 is always positive
of course we are pretending we have real a and real b

- anonymous

Thank you very much!

- myininaya

but it seems like the question you asking now is different from the one you are asking earlier
\[f'(x)=(x-a)(x-b)^2\]
we want f'>0
\[(x-a)(x-b)^2>0\]
f' is zero when x=a or x=b
|-----|---------|----
0 a b
chose number btw 0 and a like a/2
we want the following to be greater than 0
\[f'(\frac{a}{2})=(\frac{-a}{2})(\frac{a}{2}-b)^2 \]
we want to choose a number btw a and b now like (a+b)/2
again we want the following to be greater than 0
\[f'(\frac{a+b}{2})=\frac{-a+b}{2}(\frac{a-b}{2})^2=-\frac{a-b}{2}(\frac{a-b}{2})^2=-(\frac{a-b}{2})^3\]
now we choose a number after b like b+a
we also want this to be bigger than 0
\[f'(b+a)=(b)(a)^2=a^2b\]
no now we need to think of a and b such that all of these inequalities are satisfy

- myininaya

let me know if i made a mistake somewhere

- myininaya

so we have the following three inequalities:
\[\frac{-a}{2}(\frac{a}{2}-b)^2>0, -(\frac{a-b}{2})^3>0 , a^2b>0\]

- myininaya

"For which values for a and b is f increasing?"
wait i'm confused

- anonymous

I'm sorry I'll rephrase it

- myininaya

f'=0 when x=a ,x=b

- myininaya

=> f is not increasing at x=a,x=b

- Zarkon

that is not necessarly true

- myininaya

f'=(x-a)(x-b)^2?

- myininaya

f'=0 when x=a
f'=0 when x=b

- anonymous

I'm sorry it's in Swedish but the picture might help.

##### 1 Attachment

- myininaya

zarkon i want a counterexample from you

- anonymous

Thats correct, x=a and x=b gives the zeros for the derivative

- myininaya

but we are to find values a and b so that f is increasing

- myininaya

but f'=(x-a)(x-b)^2

- myininaya

f'=0 when x=a,b

- Zarkon

\[f(x)=x^3\]

- myininaya

oh fine

- anonymous

Yes I know. And you get extra credit for pointing that out in the solution.

- myininaya

no zarkon a>0 b>0

- Zarkon

def: a function \(f\) is increasing on and interval \(I\) if for any \(x,y\in I\) with \(x

- myininaya

but a, b cannot equal zero

- Zarkon

x-a can be

- Zarkon

I was just pointing out that just because the derivative is zero does not mean that the function is not increasing.

- myininaya

right i get that part
you are correct i didn't think about x^3

- asnaseer

myininaya - as Zarkon says - f'(x)=0 could be a point of inflection. a function can be increasing/decreasing through a point of inflection

- myininaya

and other functions like x^3

- anonymous

The answers says that
(x-b)^2 will always be positive.
Because of that (x-a) determines the sign of the derivative.
x > a and x != a,b
Thats is all thats needed

- myininaya

but i wonder if i can think of one example where b>a>0

- anonymous

But the thing I couldn't solve was:
Give a good motivation why (x-b)^2 is always positive

- Zarkon

the function is increasing over b
the soluituion should just be x>a

- anonymous

It's not increasing if x=b or x=a

- Zarkon

if you fix a value of x then f is not increasing(it is just a number)...you need to look over an interval....for some open set containig b the function is increasing...even though the derivative is zero at b

- asnaseer

look at this as an example: http://www.wolframalpha.com/input/?i=y%3D%28x-1%29%28x-4%29%5E2+for+x%3D0+to+5
the function is increasing from x=4 onwards
here 4 is equivalent to point x=b

- Zarkon

i should say open interval

- Zarkon

we ar given the derivative...graph the original function

- myininaya

http://www.wolframalpha.com/input/?i=find+a+and+b+such+that+-a%5E2%2F2*%28a%2F2-b%29%5E2%3E0%2C+-%28a-b%29%5E3%2F8%3E0%2C+a%5E2b%3E0
i put in those three inequalities wolfram said there is no solution to that system

- Zarkon

this is never true for b>a>0
\[\frac{-a}{2}(\frac{a}{2}-b)^2>0\]

- anonymous

The answer doesn't require an interval. Even if x will equal a at some point, it's enough to state that at this point the derivative will be = 0

- anonymous

Thanks for your help Myininaya

- Zarkon

you gave an interval (2 intervals) as an answer...
" x > a and x != a,b Thats is all thats needed"

- anonymous

Sorry sorry, I didn't think realize that. But that's the correct answer.

- Zarkon

it is not

- asnaseer

myininaya - your original inequalities were correct - they lead to the correct solution of x > a

- myininaya

but -a/2<0 since a>0

- myininaya

i don't see a solution to the problem

- asnaseer

you got this for x < a:\[f'(\frac{a}{2})=(\frac{-a}{2})(\frac{a}{2}-b)^2\]which is always negative, so function cannot be increasing over this region.
for x>a and x

**a. and finally for x > b you got:\[f'(b+a)=a^2b\]which is always positive since 0 < a < b**

**
**

- asnaseer

therefore function is increasing for x > a

- myininaya

but want f'>0 always
we wanted f to be always increasing and that gets messed up on my f'(a/2)

- myininaya

oh we wanted to know where f is increasing?

- myininaya

or did we want to choose a and b so that f is increasing everywhere?

- myininaya

first problem is x>a
but second problem has no solution

- Zarkon

it is not possible to pick an a and b such that f is increasing everywhere

- anonymous

This is an analyzing problem. We should find when f is increasing and when it's zero

- anonymous

We don't have to do that

- anonymous

Just state that if x > a, f is increasing. as long as it isn't = a or b

- anonymous

But since it can't be > a and = a.
x > a and x != b is the solution given in the answer sheet.

- asnaseer

I think your answer sheet is wrong, it should just be x > a as Zarkon stated earlier

- anonymous

if x=b then f' = 0 which means not increasing

- asnaseer

|dw:1328403876728:dw|

- asnaseer

the function can still be increasing through a point of inflection

- anonymous

Not at the point

- Zarkon

points do not increase and decrease...functions do over intervals

- asnaseer

if you pick any specific point, then you cannot say that function is increasing/decreasing, it just has some value there. increasing/decreasing applies to an interval.

- anonymous

"The rate of change at a particular moment. Same as the value of the derivative at a particular point."

- anonymous

I'm sorry I understand what you mean

- anonymous

talk about killing an ant with an h bomb

- anonymous

not that i read the whole thing, but isn't is just asking for which values of x is
\[x-a>0\] namely
\[x>a\]? if the function is continuous (and we assume it is since we assume it has a derivative everywhere) then the fact that the derivative may be postitive, then zero then positive means the funcion is increasing over the whole interval

- myininaya

yes satellite

- anonymous

oh i see now that i read the posts

- anonymous

there was some confusion over what happens at b
which reminds me

**
**

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