Two Functions intersect and have equal derivatives at x=a and x=b

[Jay520]

Sorry, wasn't sure how to describe the problem in the title.

Homework Statement

Okay, this problem has really been bugging me for a while. It's a question that I thought of when I was daydreaming in class, but now I can't stop thinking about it. I've been going crazy for days Here is the problem:

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Consider the following information to determine the answer to the question afterward.

• f (0) and g (0) = 0
• f '(0) and g '(0) = 1
• f (h) and g (h) = 1
• f '(h) and g '(h) = 0
• h is positive constant greater than 1
• f (x) and g (x) are always concave down on the interval (0, h)
• f (x) and g (x) are both continuous on the interval (0, h)
• f (x) and g (x) are both always differentiable on the interval (0, h)

On the interval (0,h), do the graphs of f (x) and g (x) have to be identical?

If not, give the algebraic expression of two functions that satisfy the above conditions and also have nonidentical graphs on the interval (0,h).

The attempt at a solution

I don't believe two functions necessarily have to be identical on the interval with those conditions. However, I can't find two functions to give an example

My first attempt was to use f(x) = sinx as one of the functions, since it satisfies these conditions when h=pi/4.

when f(x) = sinx, and h=pi/4

• f (0) = 0
• f '(0) = 1
• f (pi/4) = 1
• f '(pi/4) = 0

Now I just needed to find a function g(x) with the same conditions, but has a curve nonidentical to sinx on the interval (0, pi/4). I figured if I manipulate the semi-circle equation, I can obtain a graph with the above characteristics. So I set g(x) = a√(b-cx^2) + d, which is a transformation of the semi-circle equation.

However, after hours upon hours of attempts, I cannot find the values of a, b, c & d to satisfy all three conditions! For example, I can easily get g(0) = 0, g'(0) = 1, and g (pi/4) = 1, but then g '(pi/4) would equal something other than 0. What is happening?

I also tried other functions like the cubic function, function raised to the 1/3 power, etc. but nothing works. There's always one condition unfulfilled.

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Am I wrong in my thinking that two functions with the given conditions are not necessarily identical? I asked my GSI and he gave me some philosophical nonsense and told me I can figure it out on my own (he's obviously wrong).

PS: Could you explain it in a way that a first-semester Calculus student can understand?

 

[jedishrfu]

Consider a sin curve vs a square wave with the same periodicity at specific points the value and slope might agree but the functions are clearly different.

 

[pasmith]

Sorry, wasn't sure how to describe the problem in the title.

Homework Statement

Okay, this problem has really been bugging me for a while. It's a question that I thought of when I was daydreaming in class, but now I can't stop thinking about it. I've been going crazy for days Here is the problem:

-

Consider the following information to determine the answer to the question afterward.

• f (0) and g (0) = 0
• f '(0) and g '(0) = 1
• f (h) and g (h) = 1
• f '(h) and g '(h) = 0
• h is positive constant greater than 1
• f (x) and g (x) are always concave down on the interval (0, h)
• f (x) and g (x) are both continuous on the interval (0, h)
• f (x) and g (x) are both always differentiable on the interval (0, h)

On the interval (0,h), do the graphs of f (x) and g (x) have to be identical?

Suppose there exists a function $F: [0,h] \to \mathbb{R}$, which satisfies the following conditions:

• $F$ is continuous on $[0,h]$ and differentiable on $(0,h)$.
• $F$ is concave down on $(0,h)$.
• $F(0) = F'(0) = F(h) = F'(h) = 0$.
• There exists $x \in (0,h)$ such that $F(x) \neq 0$.

Now suppose $f : [0,h] \to \mathbb{R}$ satisfies the following:

• $f(0) = 0$, $f'(0) = 1$, $f(h) = 1$ and $f'(h) = 0$.
• $f$ is concave down on $(0,h)$.
• $f$ is continuous on $[0,h]$ and differentiable on $(0,h)$.

Then $f$ and $g = f + F \neq f$ will satisfy the conditions you have imposed.

All you need do is find such an $F$.

 

[Jay520]

Consider a sin curve vs a square wave with the same periodicity at specific points the value and slope might agree but the functions are clearly different.

Yeah, that's why I don't think f(x) and g(x) have to be equal. But I can't think of any counterexamples to the proposed question. Can you?

 

[Jay520]

Suppose there exists a function $F: [0,h] \to \mathbb{R}$, which satisfies the following conditions:

• $F$ is continuous on $[0,h]$ and differentiable on $(0,h)$.
• $F$ is concave down on $(0,h)$.
• $F(0) = F'(0) = F(h) = F'(h) = 0$.
• There exists $x \in (0,h)$ such that $F(x) \neq 0$.

Now suppose $f : [0,h] \to \mathbb{R}$ satisfies the following:

• $f(0) = 0$, $f'(0) = 1$, $f(h) = 1$ and $f'(h) = 0$.
• $f$ is concave down on $(0,h)$.
• $f$ is continuous on $[0,h]$ and differentiable on $(0,h)$.

Then $f$ and $g = f + F \neq f$ will satisfy the conditions you have imposed.

Such a function $F$ exists; for example the solution of the differential equation
$$\frac{d^4 F}{dx^4} = -1$$
subject to $F(0) = F'(0) = F(h) = F'(h) = 0$.

hmm, I'm a first-semester Calculus student. I assumed the question could be answered using some relatively basic calculus. I've never seen some of the symbols you just used.

I doubt understand your third bullet where you put "F(0) = F'(0) = F(h) = F'(h) = 0." How did you figure that out? Also, for the fourth bullet, I have no idea what that "E" looking character is.

Sorry for the hassle, but could you explain it in a way that a first-semester Calculus student can understand? I may be a few courses away from understanding your response completely.

 

[LCKurtz]

@Jay520 re your OP: It's easy to draw a few such curves:

 

[pasmith]

hmm, I'm a first-semester Calculus student. I assumed the question could be answered using some relatively basic calculus. I've never seen some of the symbols you just used.

I doubt understand your third bullet where you put "F(0) = F'(0) = F(h) = F'(h) = 0." How did you figure that out?

I want to show that if f and g satisfy the given conditions, then they need not be equal. Thus I want to write $F = g - f$, and show that $F$ doesn't have to be zero everywhere.

Since $f(0) = g(0)$, $f'(0) = g'(0)$, $f(h) = g(h)$ and $f'(h) = g'(h)$ I therefore have $F(0) = F'(0) = F(h) = F'(h) = 0$.

Also, for the fourth bullet, I have no idea what that "E" looking character is.

"$a \in A$" means "$a$ is a member of the set $A$".