Finding the Range of f(x) on [0,1]: A Challenging Calculus Problem

[utkarshakash]

Homework Statement

Let f:[0,1]→R be a function. Suppose the function f is twice differentiable, f(0)=f(1)=0 and satisfies f"(x)-2f'(x)+f(x)≥e^x, x:[0,1]. Then find range of f(x).

The Attempt at a Solution

Strictly speaking, this has been one of the toughest problems I've ever encountered throughout my course in Calculus. So, obviously I don't have an idea where to start with. But atleast, from the given information, I can say that f'(x) must be zero somewhere in [0,1] according to Rolle's Theorem. Since the inequation also involves a second order derivative, finding a solution of this differential inequation (never heard of diff. inequation btw) is beyond my reach.

 

[Mark44]

Homework Statement

Let f:[0,1]→R be a function. Suppose the function f is twice differentiable, f(0)=f(1)=0 and satisfies f"(x)-2f'(x)+f(x)≥e^x, x:[0,1]. Then find range of f(x).

The Attempt at a Solution

Strictly speaking, this has been one of the toughest problems I've ever encountered throughout my course in Calculus. So, obviously I don't have an idea where to start with. But atleast, from the given information, I can say that f'(x) must be zero somewhere in [0,1] according to Rolle's Theorem. Since the inequation also involves a second order derivative, finding a solution of this differential inequation (never heard of diff. inequation btw) is beyond my reach.

If you turn your inequality into an equation, can you solve that?

I did that, and got a solution that consists of a quadratic times e²; i.e., y(x) = (Ax² + Bx + C)e^x. The quadratic is a parabola that opens up and has x-intercepts at 0 and 1. My thinking is that if f(x) is any twice-differentiable function that is at least as large as y(x), and intersects the x-axis at 0 and 1, then that will be a solution of your differential inequality.

I'm not 100% confident in my reasoning here - I haven't ever seen differential inequalities, either. If the direction I took makes sense, you could at least put a lower bound on the range.

 

[haruspex]

I'm not sure what the question is asking. Clearly f is not fully defined, so different functions satisfying the condition might have different ranges. Is it the union of the ranges that's required? The set of ranges?

 

[crownedbishop]

Well, the range is just R. No? Unless you are talking about the image of f. By the way, I do not believe that f is uniquely defined, as haruspex pointed out.

 

[PeroK]

Hint: Let $g(x) = f(x)e^{-x}$

 

[haruspex]

Well, the range is just R. No?

Why? I think it unlikely that arbitrarily negative values can be reached.
If you take the inequality as exact, as Mark44 did, and plug in the known facts, you'll find 2a >= 1, b = -a and c = 0. You can then find a local min in the interval as a function of a.

 

[PeroK]

Let $g(x) = f(x)e^{-x}$

g satisfies g(0) = g(1) = 0 and g''(x) ≥ 1

Let $g_n(x) = nx(x-1)$

So: $f_n(x) = nx(x-1)e^x$

So, the range of f is <= 0 and can be arbitrarily large and -ve.

 

[utkarshakash]

I'm not sure what the question is asking. Clearly f is not fully defined, so different functions satisfying the condition might have different ranges. Is it the union of the ranges that's required? The set of ranges?

Ah, thanks for pointing out. The question actually gives me 4 options and asks me to select a correct one. Here are those 4 options:-

Q. Which of the following is true for 0<x<1?
A) 0<f(x)<∞
B)-1/2 < f(x)< 1/2
C)-1/4<f(x)<1
D)-∞<f(x)<0

 

[utkarshakash]

Did anyone get the answer?

 

[haruspex]

Did anyone get the answer?

PeroK's analysis convinced me.

 

[HallsofIvy]

Well, the range is just R. No? Unless you are talking about the image of f. By the way, I do not believe that f is uniquely defined, as haruspex pointed out.

Why? I think it unlikely that arbitrarily negative values can be reached.
If you take the inequality as exact, as Mark44 did, and plug in the known facts, you'll find 2a >= 1, b = -a and c = 0. You can then find a local min in the interval as a function of a.

You two are using different definitions of "range". crownedbishop is saying that if f is "from A to B" then its range is "B", even if f(A), the "image" of f (or, better, the "image of A under f")is a proper subset of B.

 

[haruspex]

crownedbishop is saying that if f is "from A to B" then its range is "B", even if f(A), the "image" of f (or, better, the "image of A under f")is a proper subset of B.

True, range can mean either codomain or image of domain. But it seems clear that image of domain is what's wanted here.

 

[utkarshakash]

Let $g(x) = f(x)e^{-x}$

g satisfies g(0) = g(1) = 0 and g''(x) ≥ 1

Let $g_n(x) = nx(x-1)$

So: $f_n(x) = nx(x-1)e^x$

So, the range of f is <= 0 and can be arbitrarily large and -ve.

Your approach is a perfect one. May I know how did it occur to you to assume another function g(x)=f(x)e^-x? Were you led by your intuition or just a lucky guess? How exactly did you figure it out? Were you able to realize it quickly or did it took a while?

 

[PeroK]

Your approach is a perfect one. May I know how did it occur to you to assume another function g(x)=f(x)e^-x? Were you led by your intuition or just a lucky guess? How exactly did you figure it out? Were you able to realize it quickly or did it took a while?

I just saw the pattern of f'' - 2f' + f and thought I might be able to find a function whose second derivative that was. I used what's called an "integrating factor", a technique used in solving of differential equations:

Since you ask, it was the first thing I thought of when I saw the question.