anonymous
  • anonymous
Prove that x^2 is always positive. I understand why, but I don't know how to write out.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Mertsj
  • Mertsj
Try an indirect proof....assume that it is negative and show that it leads to a contradiction.
nenadmatematika
  • nenadmatematika
for every x in R y=x^2 stays equal to zero or bigger than zero
myininaya
  • myininaya
what do you mean you don't know how to write x^2>0?

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myininaya
  • myininaya
and yes because one counterexample for all complex would be if x=i i^2=-1<0
nenadmatematika
  • nenadmatematika
correct....just look out because it could be equal to zero when x=0
myininaya
  • myininaya
yep yep
myininaya
  • myininaya
so the best thing to say is x^2>=0 for all reals
Zarkon
  • Zarkon
show thats its minumm value is 0
myininaya
  • myininaya
which nenad already said
Zarkon
  • Zarkon
I butchered the crap out of that word....minimum ;)
myininaya
  • myininaya
zarkon you don't have to know how to spell
myininaya
  • myininaya
you just have to get it a little right
Zarkon
  • Zarkon
good...because I can't
anonymous
  • 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
  • myininaya
f'=2x=0 => x=0 which means x=0 is a critical number f''=2 which means it is concave up
myininaya
  • 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
  • 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
  • anonymous
Thank you very much!
myininaya
  • 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
  • myininaya
let me know if i made a mistake somewhere
myininaya
  • 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
  • myininaya
"For which values for a and b is f increasing?" wait i'm confused
anonymous
  • anonymous
I'm sorry I'll rephrase it
myininaya
  • myininaya
f'=0 when x=a ,x=b
myininaya
  • myininaya
=> f is not increasing at x=a,x=b
Zarkon
  • Zarkon
that is not necessarly true
myininaya
  • myininaya
f'=(x-a)(x-b)^2?
myininaya
  • myininaya
f'=0 when x=a f'=0 when x=b
anonymous
  • anonymous
I'm sorry it's in Swedish but the picture might help.
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myininaya
  • myininaya
zarkon i want a counterexample from you
anonymous
  • anonymous
Thats correct, x=a and x=b gives the zeros for the derivative
myininaya
  • myininaya
but we are to find values a and b so that f is increasing
myininaya
  • myininaya
but f'=(x-a)(x-b)^2
myininaya
  • myininaya
f'=0 when x=a,b
Zarkon
  • Zarkon
\[f(x)=x^3\]
myininaya
  • myininaya
oh fine
anonymous
  • anonymous
Yes I know. And you get extra credit for pointing that out in the solution.
myininaya
  • myininaya
no zarkon a>0 b>0
myininaya
  • myininaya
0
Zarkon
  • Zarkon
def: a function \(f\) is increasing on and interval \(I\) if for any \(x,y\in I\) with \(x
myininaya
  • myininaya
but a, b cannot equal zero
Zarkon
  • Zarkon
x-a can be
Zarkon
  • Zarkon
I was just pointing out that just because the derivative is zero does not mean that the function is not increasing.
myininaya
  • myininaya
right i get that part you are correct i didn't think about x^3
asnaseer
  • 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
  • myininaya
and other functions like x^3
anonymous
  • 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
  • myininaya
but i wonder if i can think of one example where b>a>0
anonymous
  • anonymous
But the thing I couldn't solve was: Give a good motivation why (x-b)^2 is always positive
Zarkon
  • Zarkon
the function is increasing over b the soluituion should just be x>a
anonymous
  • anonymous
It's not increasing if x=b or x=a
Zarkon
  • 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
  • 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
  • Zarkon
i should say open interval
Zarkon
  • Zarkon
we ar given the derivative...graph the original function
myininaya
  • 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
  • Zarkon
this is never true for b>a>0 \[\frac{-a}{2}(\frac{a}{2}-b)^2>0\]
anonymous
  • 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
  • anonymous
Thanks for your help Myininaya
Zarkon
  • Zarkon
you gave an interval (2 intervals) as an answer... " x > a and x != a,b Thats is all thats needed"
anonymous
  • anonymous
Sorry sorry, I didn't think realize that. But that's the correct answer.
Zarkon
  • Zarkon
it is not
asnaseer
  • asnaseer
myininaya - your original inequalities were correct - they lead to the correct solution of x > a
myininaya
  • myininaya
but -a/2<0 since a>0
myininaya
  • myininaya
i don't see a solution to the problem
asnaseer
  • 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
  • asnaseer
therefore function is increasing for x > a
myininaya
  • myininaya
but want f'>0 always we wanted f to be always increasing and that gets messed up on my f'(a/2)
myininaya
  • myininaya
oh we wanted to know where f is increasing?
myininaya
  • myininaya
or did we want to choose a and b so that f is increasing everywhere?
myininaya
  • myininaya
first problem is x>a but second problem has no solution
Zarkon
  • Zarkon
it is not possible to pick an a and b such that f is increasing everywhere
anonymous
  • anonymous
This is an analyzing problem. We should find when f is increasing and when it's zero
anonymous
  • anonymous
We don't have to do that
anonymous
  • anonymous
Just state that if x > a, f is increasing. as long as it isn't = a or b
anonymous
  • anonymous
But since it can't be > a and = a. x > a and x != b is the solution given in the answer sheet.
asnaseer
  • asnaseer
I think your answer sheet is wrong, it should just be x > a as Zarkon stated earlier
anonymous
  • anonymous
if x=b then f' = 0 which means not increasing
asnaseer
  • asnaseer
|dw:1328403876728:dw|
asnaseer
  • asnaseer
the function can still be increasing through a point of inflection
anonymous
  • anonymous
Not at the point
Zarkon
  • Zarkon
points do not increase and decrease...functions do over intervals
asnaseer
  • 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
  • anonymous
"The rate of change at a particular moment. Same as the value of the derivative at a particular point."
anonymous
  • anonymous
I'm sorry I understand what you mean
anonymous
  • anonymous
talk about killing an ant with an h bomb
anonymous
  • 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
  • myininaya
yes satellite
anonymous
  • anonymous
oh i see now that i read the posts
anonymous
  • anonymous
there was some confusion over what happens at b which reminds me

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