Min/max functions in single-variable calc

[FatCat0]

[SOLVED] Min/max functions in single-variable calc

1. Homework Statement
7. If x is positive and m greater than one, prove that x^m - 1 - m(x-1) is not negative using max-min.




2. Homework Equations



3. The Attempt at a Solution

I can't figure out a secondary equation to get the original in terms of x or m so I can do the first and second derivatives. There must be something stupid I'm missing... I don't know if you help in the situations where the first step is needed

 

[D H]

With $f(x)=x^m-1-m(x-1)$, where does f(x) reach extremum?

 

[FatCat0]

At m = 1, f(x) = 0 regardless. When m is any greater than that f(x) is greater than 0 (since x^m increases exponentially whereas m(x-1) increases linearly). But I don't know how to show this without simply giving data; it has to be mathematical.

 

[FatCat0]

I guess my problem is more in conveying the information to my teacher than actually solving the problem...

 

[D H]

You didn't answer my question. What level math are you in? Do you know about using derivatives to find extrema?

 

[FatCat0]

Calc BC (high school level), and yes. I figured out what I was doing wrong; m is a constant and can be treated as thus. Doing the limit that you suggested before editing that post was also a bit misleading.

Since m is constant, you can take the derivative of the original equation and get f'(x) = mx^(m-1) - m. Set it equal to 0, you end up with mx^(m-1)=m, them x^(m-1) = 1.

Solve for x, you get x^m - x = 0, or x^m = x.

Since m > 1 and thus cannot be 1, and x > 0 and cannot be 0, x must = 1.

Go back to the original equation f(x) = 1^m - 1 - m(x - 1)

Simplify to 1^m - 1. For any value of m, 1^m will always equal 1, thus the equation always = 0 which isn't negative.

Which begs the question, ARE there any extrema in this equation? It's always 0...

Edit: Oh, and thanks as well. I did have my "Oh, duh!" moment while looking at this post, so...

 

[D H]

Which begs the question, ARE there any extrema in this equation? It's always 0...

The derivative is always zero because you forced it to be so by choosing x=1. The original function, $f(x)=x^m-1-m(x-1)$ is not identically zero. For example, $f(0)=m-1$. The derivative is zero at x=1 for all m. This does not mean the derivative is identically zero for all x.

Do you know about the second derivative test to determine whether an extremum is a minimum or a maximum?

 

[FatCat0]

Yes, I do. if f '' (x) > 0, x is a min, opposite it's a max, and if f '' (x) = 0 it's a Point of Inflection (usually anyway, for some reason (something that came up earlier in the assignment) in a cosine function my min had an f '' (x) of 0).

I'm not catching something. The second derivative would be (m-1)(m)x^(m-2). If m > 1 and x > 0, then the second derivative must be (+) (+) (+)^(+/-), overall becoming positive, simply proving that this is a minimum. But I don't see how one can go from there to prove that that minimum is greater than 0 in f(x) without "forcing" it to be 0.

 

[FatCat0]

Oh, is that really forcing the equation to be 0? We're supposed to work with critical numbers, and the only one (that I found anyway) was x = 1, given all of the criteria.

I'm really sorry if I'm like making you bash your head into your desk for not noticing something painfully obvious.

 

[D H]

First, a minor correction on post #6.

Since m is constant, you can take the derivative of the original equation and get f'(x) = mx^(m-1) - m.

Correct.

Set it equal to 0, you end up with mx^(m-1)=m, them x^(m-1) = 1.

Correct.

Solve for x, you get x^m - x = 0, or x^m = x.

Why did you do this? Doing this introduces zero as a solution, which it isn't since $m>1$. One is obviously a solution to $x^{m-1}=1$. The only other possible real root to the expression $x^{m-1}=1$ is $x=-1$, and you can ignore this possibility since $x>0$.

if f '' (x) > 0, x is a min, opposite it's a max, and if f '' (x) = 0 it's a Point of Inflection (usually anyway, for some reason (something that came up earlier in the assignment) in a cosine function my min had an f '' (x) of 0).

I'm not catching something. The second derivative would be (m-1)(m)x^(m-2). If m > 1 and x > 0, then the second derivative must be (+) (+) (+)^(+/-), overall becoming positive, simply proving that this is a minimum.

To summarize, this is what you have deduced

• The function has one and only one extremum for positive x, and this is at x=1 (post #6)
• This one extremum is a minimum (post #8).

You need just one more piece of information to finish the problem. Hint: What is the minimum value of the function?

 

[FatCat0]

ze...ro? Just prove that the minimum of this function for positive x, x = 1, is 0 which proves that everything else is greater than zero, ergo positive? Didn't I do that in post 6?

"Simplify to 1^m - 1. For any value of m, 1^m will always equal 1, thus the equation always = 0 which isn't negative."

I might have worded it incorrectly, I should have stated that the minimum is 0, but is that the last piece of needed info?

And on the introducing 0 as an option, x can't equal 0 because x is positive (>0). I...don't know why I did it that way either, to tell you the truth. It was just easier to look at I think.

 

[D H]

You have shown the minimum value the function can obtain for any positive x is zero. Now look at the problem statement itself: "If x is positive and m greater than one, prove that x^m - 1 - m(x-1) is not negative". There is only one extremum, and it is a minimum. This minimum is zero. How can the function be negative?

Suppose the function does have a negative value for some positive x, call it x₀. This means the function must have a local maximum somewhere between x=1 and x=x₀. But it doesn't because the function has only one extremum.

 

[FatCat0]

Alright. Danke Schoen D H.

 

[D H]

You're welcome.