Let f:(a,b)→R be a differentiable map and let x0∈(a,b) be a critical point. Prove that if there exists δ&gt;0 such that f′(x)≥0 for all x∈(x0−δ,x0) and f′(x)≤0 for all x∈(x0,x0+δ) then x0 is a local maximum of f.

Question

Let f:(a,b)→R be a differentiable map and let x0∈(a,b) be a critical point. Prove that if there exists δ&amp;gt;0 such that f′(x)≥0 for all x∈(x0−δ,x0) and f′(x)≤0 for all x∈(x0,x0+δ) then x0 is a local maximum of f.

Mason Hall · Accepted Answer

Step 1  Suppose x0 is not a local maximum.   Then we should be able to find x such that |x−x0|&amp;lt;δ and f(x)&amp;gt;f(x0).  Now apply the mean value theorem to relate the difference f(x)−f(x0) to the derivative of f evaluated at some point near x0, and use what you know about the sign of the derivative near x0.

Virginia Palmer · Accepted Answer

Step 1We claim thatf(x)≥f(x0)for everyx∈(x0−δ,x0+δ)which proves that x0 is a local minimum of f.Pick an arbitrary x∈(x0−δ,x0+δ)If x=x0, then f(x)=f(x0), so we have equality in this case. If x≠x0, then either x∈(x0−δ,x0)or x∈(x0,x0+δ)We consider each case separately:Case 1: x∈(x0−δ,x0)Applying the mean value theorem to f on the interval [x,x0] implies that there is a c∈(x,x0) such thatf(x0−f(x)}{x0−x}=f′(c)Since c∈(x,x0)⊂(x0−δ,x0) it follows that f′(c)≤0 and consequentlyf(x0)−f(x)x0−x≤0From x&amp;lt;x0, it follows that x0−x&amp;gt;0, so(x0−x)f(x0)−f(x)x0−x≤0⇒f(x0)−f(x)≤0⇒f(x)≥f(x0)Case 2: x∈(x0,x0+δ)Applying the mean value theorem to f on the interval [x0,x] implies that there is a c∈(x0,x) such thatf(x)−f(x0)x−x0=f′(c)This time, we have f′(c)≥0

Let f:(a,b)\rightarrow R be a differentiable map and let x_{0}\in(a,b)

Answered question

Answer & Explanation

New Questions in Analysis