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?
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
?
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
?
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.