Thread: minimum sign question

prove that if is the minimum point
then
hint: look at f(0) and f'(0).
how i tried to solve it.
f(0)=0 f'(0)=1
by limit derivative definition
so by limit definition for

what now?

If f(0) = 0, and f'(0) = 1, then what must happen just to the left of the origin?

what it has to do with how i tried to solve it?

If you suspect that f(x) < 0 for some x to the left of the origin, then posit epsilon less than 1 in magnitude, and x < 0. What does that do to your inequalities?

we are talking about the value on x0
why you say " for some x to the left of the origin"
why you say that x0<0
?

Originally Posted by transgalactic

we are talking about the value on x0
why you say " for some x to the left of the origin"
why you say that x0<0
?

Think about Ackbeet's suggestion: if f(0)= 0 and f'(0)> 0 (so that f is an increasing function there), what can you say about f(x) for x slightly less than 0?

Originally Posted by transgalactic

we are talking about the value on x0
why you say " for some x to the left of the origin"
why you say that x0<0
?

Because it's not too hard to show that f(x) > 0 for all x > 0. Think about it. Everything in sight is positive until x = pi. At that point, the x^4 completely dwarfs the sin(x), which is modulated by something that's getting smaller anyway.

Plot the function, and you'll see what I mean. The minimum happens to the left of the origin, right around -0.46.

yes i understand that for x<0 the function is negative
and that our minimum has to be less then f(0)=0 .

ok but its only words i have to show that in the equation.
if i choose epsilon =1/2

so by limit definition for
what to do next
?

Take the inequality in the OP, and multiply through by x, which is negative. What happens to the inequality signs? And then what can you say?

if we multply by negatibe x
so by limit definition for
but i cant see how from that we can conclude that the minimum is negative


i can guess that
for every negative x
f(x) is between two negative values
so it have to be negative by the intemideate value theorem?

but i dont have actual points the for the intermidieate value theorem
i have expressions of x


or does it sufficient to say that its between two negatives so its negative too.

is it a proper proof
?

Originally Posted by transgalactic

if we multply by negatibe x
so by limit definition for
but i cant see how from that we can conclude that the minimum is negative

i can guess that
for every negative x
f(x) is between two negative values
so it have to be negative by the intemideate value theorem?

but i dont have actual points the for the intermidieate value theorem
i have expressions of x

but our surriundings is -delta<x<0

not delta from both sides
correct?

Look. You have



If you choose then

.

Multiplying by an x that we assume is negative yields



But, since it follows that Hence, we can tack on the inequality to our previous one thus:

yes but in the limit definition whe have for
-delta<x<delta

but we talk only about the negative x
so only half of the section is being uses

so in the limit definition what section two write
?

So the logic works like this: since the limit exists and is 1, it is the case that for every there is a such that if then



Let Then there is a such that if then

Pick Then It follows that



And then you work out the inequalities just like before, in my post # 13, only with 's instead of x's. Finally, you can say that

since is the minimum.