A Hard "first year" Question That Involves a Few Areas of Caculus

[The Student]

Help on just one of a, b, c, d or e would be great.

Question: Suppose f is differentiable on [a,b] but f′ is not necessarily continuous, with m = f′(a) < f′(b) = M. Let v ∈ (m,M) and define g(x) = f(x) − vx.

(a) Show that g achieves a global minimum value on [a,b] at some point c ∈ [a,b].

(b) Find g′(x).

(c) Prove that g is strictly decreasing near a and strictly increasing near b.

(d) Show that the number c in part (a) cannot be equal to either a or b.

(e) Prove that f′(c) = v.

My answer:

(a) f'(a) < f'(b) only tells me that the function increases more at b than at a. How does this tell me anything about what happens between a and b? And if I don't know what else f(x) does, how could I possibly know what g(x) does?

(b) g'(x) = f'(x) - v (I don't think that this is what the question wants).

(c) g'(a) = f'(a) - va. I don't know if a is < or > than 1, so I don't know the sign of g'(a), or g'(b) for similar reasons.

(d) I don't know how we could possibly know this. I don't see any clues at all.

(e) Like (d), I just don't see any ways to figure this out.

*I wish that I could say that it might be the question that is missing information, but the professor used this as a question in a 2008 final exam, and then gave us this question for practice.

 

[pka]

Help on just one of a, b, c, d or e would be great.
Question: Suppose f is differentiable on [a,b] but f′ is not necessarily continuous, with m = f′(a) < f′(b) = M. Let v ∈ (m,M) and define g(x) = f(x) − vx.

(a) Show that g achieves a global minimum value on [a,b] at some point c ∈ [a,b].

(b) Find g′(x).

(c) Prove that g is strictly decreasing near a and strictly increasing near b.

(d) Show that the number c in part (a) cannot be equal to either a or b.

(e) Prove that f′(c) = v.

Here are some random thoughts.
We know that \(\displaystyle g\) is both continuous and differentiable on \(\displaystyle [a,b]\).
Therefore, it has a global minimum there, So \(\displaystyle f(c)-vc\) is the value of the minimum.

You correctly found \(\displaystyle g'(x)\), so \(\displaystyle g'(c)=0\) WHY? Does that answer (e)?

 

[The Student]

Here are some random thoughts.
We know that \(\displaystyle g\) is both continuous and differentiable on \(\displaystyle [a,b]\).
Therefore, it has a global minimum there, So \(\displaystyle f(c)-vc\) is the value of the minimum.

You correctly found \(\displaystyle g'(x)\), so \(\displaystyle g'(c)=0\) WHY? Does that answer (e)?

I understand now, but I only understand how to do this because of the information given in question (a). In other words, are we suppose to know that g achieves a global minimum at g(c) because the question implies it, or is this something that we could have figured out without reading questions (a) to (e)?

 

[pka]

I understand now, but I only understand how to do this because of the information given in question (a). In other words, are we suppose to know that g achieves a global minimum at g(c) because the question implies it, or is this something that we could have figured out without reading questions (a) to (e)?

I do not understand your question.
You are expected to know at a continuous function attains its maximum and its minimum on bounded closed interval.
You are expected to know that the derivative is zero at absolute minimum that is not an endpoint of the interval.

 

[The Student]

I do not understand your question.
You are expected to know at a continuous function attains its maximum and its minimum on bounded closed interval.
You are expected to know that the derivative is zero at absolute minimum that is not an endpoint of the interval.

Oh thank-you so much!!!! You are amazing!!!!