1. ## minimum sign question

$f(x)=x^{4}+(sinx)e^{-x^{2}}$
prove that if $x_{0}$ is the minimum point
then $f(x_{0})<0$
hint: look at f(0) and f'(0).
how i tried to solve it.
f(0)=0 f'(0)=1
by limit derivative definition $f'(0)=lim_{x->0}\frac{f(0+x)-f(0)}{x-0}=lim_{x->0}\frac{f(x)}{x}=1$
so by limit definition for $-\delta
$1-\epsilon<\frac{f(x)}{x}<1+\epsilon$
what now?

2. ## Re: minimum sign question

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

3. ## Re: minimum sign question

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

4. ## Re: minimum sign question

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?

5. ## Re: minimum sign question

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
?

6. ## Re: minimum sign question

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?

7. ## Re: minimum sign question

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.

8. ## Re: minimum sign question

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 $-\delta $0.5<\frac{f(x)}{x}<1.5$
what to do next
?

9. ## Re: minimum sign question

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?

10. ## Re: minimum sign question

if we multply by negatibe x
so by limit definition for $-\delta $0.5x>f(x)>1.5x$
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
?

11. ## Re: minimum sign question

Originally Posted by transgalactic
if we multply by negatibe x
so by limit definition for $-\delta $0.5x>f(x)>1.5x$
but i cant see how from that we can conclude that the minimum is negative
$0>0.5x>f(x)>1.5x.$

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

12. ## Re: minimum sign question

but our surriundings is -delta<x<0

not delta from both sides
correct?

13. ## Re: minimum sign question

Look. You have

$1-\epsilon<\frac{f(x)}{x}<1+\epsilon.$

If you choose $\epsilon=0.5,$ then

$0.5<\frac{f(x)}{x}<1.5$.

Multiplying by an x that we assume is negative yields

$0.5x>f(x)>1.5x.$

But, since $x<0,$ it follows that $0.5x<0.$ Hence, we can tack on the inequality to our previous one thus:

$0>0.5x>f(x)>1.5x.$

14. ## Re: minimum sign question

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
?

15. ## Re: minimum sign question

So the logic works like this: since the limit exists and is 1, it is the case that for every $\epsilon>0,$ there is a $\delta>0$ such that if $|x|<\delta,$ then

$\left|\frac{f(x)-f(0)}{x-0}-1\right|<\epsilon.$

Let $\epsilon=0.5.$ Then there is a $\delta_{0.5}>0$ such that if $|x|<\delta_{0.5},$ then $|(f(x)/x)-1|<0.5.$

Pick $x_{1}\in(-\delta_{0.5},0).$ Then $|x_{1}|<\delta_{0.5}.$ It follows that

$|(f(x_{1})/x_{1})-1|<0.5.$

And then you work out the inequalities just like before, in my post # 13, only with $x_{1}$'s instead of x's. Finally, you can say that

$f(x_{0}) since $x_{0}$ is the minimum.

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