Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

yep yep

Answer 7

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

Answer 8

show thats its minumm value is 0

Answer 9

which nenad already said

Answer 10

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

Answer 11

zarkon you don't have to know how to spell

Answer 12

you just have to get it a little right

Answer 13

good...because I can't

Answer 14

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?

Answer 15

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

Answer 16

so a and b are critical numbers

\[f''=(x-b)^2+(x-a) \cdot 2(x-b)=(x-b)[x-b+2(x-a))]=(x-b)[3x-b-2a]\]

\[f''(a)=(a-b)[3a-b-2a]=(a-b)[a-b]=(a-b)^2=(b-a)^2>0\]

\[f''(b)=0 \text{ inconclusive }\]

Answer 17

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

Answer 18

Thank you very much!

Answer 19

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

Answer 20

let me know if i made a mistake somewhere

Answer 21

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

Answer 22

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

Answer 23

I'm sorry I'll rephrase it

Answer 24

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

Answer 25

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

Answer 26

that is not necessarly true

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

zarkon i want a counterexample from you

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

f'=0 when x=a,b

Answer 35

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

Answer 36

oh fine

Answer 37

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

Answer 38

no zarkon a>0 b>0

Answer 39

0<a<b

Answer 40

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



\[x^3\] satisfies this definition on\(\mathbb{R}\)

Answer 41

but a, b cannot equal zero

Answer 42

x-a can be

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

and other functions like x^3

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

the function is increasing over b

the soluituion should just be x>a

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

i should say open interval

Answer 55

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

Answer 56

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

Answer 57

this is never true for b>a>0

\[\frac{-a}{2}(\frac{a}{2}-b)^2>0\]

Answer 58

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

Answer 59

Thanks for your help Myininaya

Answer 60

you gave an interval (2 intervals) as an answer...

" x > a and x != a,b Thats is all thats needed"

Answer 61

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

Answer 62

it is not

Answer 63

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

Answer 64

but -a/2<0 since a>0

Answer 65

i don't see a solution to the problem

Answer 66

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<b you got:\[f'(\frac{a+b}{2})=-(\frac{a-b}{2})^3=-\frac{a-b}{2}(\frac{a-b}{2})^2=\frac{b-a}{2}(\frac{a-b}{2})^2\]which is always positive since b > a.
and finally for x > b you got:\[f'(b+a)=a^2b\]which is always positive since 0 < a < b

Answer 67

therefore function is increasing for x > a

Answer 68

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

Answer 69

oh we wanted to know where f is increasing?

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

We don't have to do that

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

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