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lep11

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## Homework Statement

Let f: ℝ→ℝ be continuous function such that f(x) approaches 0 when x---> -∞ and f(0)=2. In addition, f is decreasing when x≥3. Prove f has a maximum (greatest value) in ℝ.

## Homework Equations

Theorem: If function f is continuous on a closed interval [a, b], then f has maximum and minimum in [a, b].

3. The Attempt at a Solution

3. The Attempt at a Solution

Because f(x) approaches 0 when x ---> -∞ then there exists a<0 such that a<0 ⇒ |f(x)-0|=|f(x)|<2. So a<0 ⇒ f(x)<2. Because f is continuous in ℝ, it's also continuous on a closed interval [a,3]⊂ℝ and there's the theorem in our lecture notes that says now ∃x

_{0}∈[a,3] such that f(x)≤f(x

_{0}) ∀x∈[a,3]. Since a is negative, 0∈[a,3] and therefore f(x

_{0})≥f(0)=2.

*Because f is decreasing when x≥3, f(x)≤f(x*_{0}), when x>x_{0}≥3.The idea is to divide ℝ into intervals so that we get closed interval to apply the theorem.

In a summary:

##a<0 ⇒ f(x)<2≤f(x_0)##

##x∈[a,3] ⇒f(x)≤f(x_0)##

##x≥3 ⇒f(x)≤f(x_0)##

∴ f has greatest value/maxima in ℝ.

Is this proof correct and rigorious enough? I think I have made a mistake in the end when it comes to the fact that f is decreasing when x≥3.

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