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Prove that if $f(x)$ has a least upper bound $M$ then for every $\epsilon\gt 0$ there exists an $\bar{x}$ in the domain of $f(x)$ such that $|f(\bar{x})-M|\lt\epsilon$ .

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Lesson Parent:

Geometry of Functions II: The Extreme-Value Theorem

Problem:

Answer:

Solution:

Since $M$ is an upper bound then for every $x$ in the domain of $f(x)$ then <Sign in to see all the formulas>.

Let's denote $n$ as any positive integer. Now since $M$ is also the least upper bound then $M-1/n$ cannot be an upper bound. Since $M-1/n$ is not an upper bound, there exists <Sign in to see all the formulas> in the domain of $f(x)$ such that <Sign in to see all the formulas>.

Therefore, for every positive integer $n$ there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>.

Now let's do some quick algebra:

<Sign in to see all the formulas>

Therefore, <Sign in to see all the formulas>.

Now we know that for every <Sign in to see all the formulas> there exists a natural number $n$ such that <Sign in to see all the formulas>. So now we know that for every <Sign in to see all the formulas> there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>.

Prove that if f(x)f(x) has a least upper bound MM then for every ϵ>0\epsilon\gt 0 there exists an ˉx\bar{x} in the domain of f(x)f(x) such that |f(ˉx)−M|<ϵ|f(\bar{x})-M|\lt\epsilon.

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SAMPLE SOLUTION

<Sign in to see all the formulas>

GAIN AN ADVANTAGE

Prove that if $f(x)$ has a least upper bound $M$ then for every $\epsilon\gt 0$ there exists an $\bar{x}$ in the domain of $f(x)$ such that $|f(\bar{x})-M|\lt\epsilon$ .