Suppose that f is a differentiable real function in

[Jamin2112]

Homework Statement

Suppose that f is a differentiable real function in an open set E (which is a subset of) ℝⁿ, and that that f has a local maximum at a point x in E. Prove that f'(x)=0.

Homework Equations

Definition. Suppose E is an open set in ℝⁿ, f maps E into ℝ^m, and x is an element of E. If there exists a linear transformation A of ℝⁿ into ℝ^m such that

lim_h→0 |f(x+h)-f(x)-Ah|/|h|=0,
​

then f is differentiable at x, and we write

f'(x)=A.​

The Attempt at a Solution

f has a local max at x, so there exists a δ>0 such that f(y)≤f(y) for all y at which |y-x|<δ. Let h=x+y. Whenever |h|≤δ we have f(x)≥f(x+h) or f(x)≥f(x-h).

Now I need to use some algebraic cut-and-paste wizardry to end up with |f(x+h)-f(x)|/|h|. Could I please get an inconspicuous suggestion from the peanut gallery here?

 

[jake929]

your problem statement includes f'(x)=0 your relavent equation has a similar form of f'(x)=A, if you show that A = 0 at a local maxima then you should have your answer. Hint: Make sure that you understand the physical meaning of a local maxima.

 

[Jamin2112]

your problem statement includes f'(x)=0 your relavent equation has a similar form of f'(x)=A, if you show that A = 0 at a local maxima then you should have your answer. Hint: Make sure that you understand the physical meaning of a local maxima.

Did not the first sentence of my attempt capture the concept of a local maxima?

 

[jake929]

I will paraphrase your problem statement and maybe that will help you understand what I mean. The problem is asking you to prove that the slope of a line at a local maxima is 0 using the fundamental theorem of calculus. This is something that will be assumed for later proofs. If you look at the fundamental theorem it will show an equation that looks like what you have seen in pre-calculus classes for calculating the slope (y2-y1)/(x2-x1) = slope. Try and show that the fundamental theorem yields a slope of zero at the maxima. Post your work if you want further help.

 

[HallsofIvy]

jake929, what you are suggesting is valid only in one dimension.

 

[jake929]

Yes, and even though the problem statement is in two dimensions the definition is in one