Proof given ##x < y < z## and a twice differentiable function

In summary, the proof discusses properties of a twice differentiable function within the context of the inequality \(x < y < z\). It likely explores the implications of the function's behavior and its derivatives in this interval, potentially focusing on aspects such as monotonicity, concavity, or the mean value theorem. The proof aims to establish conclusions about the function based on these characteristics.
  • #1
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1717374255292.png

My proof is

Since ##f'## is increasing then ##x < y <z## which then ##f(x) < f(y) < f(z)##

This is because,

##f''(t) \ge 0## for all t

## \rightarrow \int \frac{df'}{dt} dt \ge \int 0~dt = 0## for all t

##\rightarrow \int df' \geq 0## for all t
##f ' \geq 0## for all t

##\frac{df}{dt} \geq 0## for all t

##\int df \geq \int 0~dt## for all t

##f(t) \geq 0##

Now ##\frac{f(y) - f(x)}{y - x} \geq 0##

##\frac{f(z) - f(y)}{z - y}##

Assume ##y - x = z - y = c##

##\frac{f(y) - f(x)}{c} \geq 0 \implies f(y) - f(x) \geq 0##

##\frac{f(z) - f(y)}{c} \geq 0 \implies f(z) - f(y) \geq 0##

Thus we, consider two cases,

(1) ##f(z) - f(y) \geq f(y) - f(x) \geq 0##

(2) ##f(y) - f(x) \geq f(z) - f(y) \geq 0##

Note that (2) is impossible since ##f(x) < f(y) < f(z)##

##f(y) \geq 0 \implies \frac{f(y) - f(z)}{y - x} \geq \frac{0}{y - x} = 0##

##f(z) \geq 0 \implies \frac{f(z) - f(y)}{z - y} \geq \frac{0}{z - y} = 0##

We can assume that ##z - y = y - x##, since one possible function is ##f(x) = x^n## when ##n \in \mathbb{N}##. Consider case ##n = 1##, then there is a function so that ##f(z) - f(y) \geq f(y) - f(x)## however, for ##n > 1## ##f(z) - f(y) \geq f(y) - f(x)## Of course, we have only considered one case of the polynomial functions and it can be generalized to any increasing function I think.

Does anybody please know where to prove from here?

Thanks!
 
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  • #2
ChiralSuperfields said:
f′≥0 for all t
This conclusion is incorrect. Consider the counterexample: ##f=x^2##.
 
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  • #3
Hill said:
This conclusion is incorrect. Consider the counterexample: ##f=x^2##.
What do you mean? For ##x<z## we get for ##f(t)=t^2##
$$
\dfrac{f(y)-f(x)}{y-x}=\dfrac{y^2-x^2}{y-x}=y+x< z+y=\dfrac{z^2-y^2}{z-y}=\dfrac{f(z)-f(y)}{z-y}
$$
so where is the problem?
 
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  • #4
fresh_42 said:
What do you mean? For ##x<z## we get for ##f(t)=t^2##
$$
\dfrac{f(y)-f(x)}{y-x}=\dfrac{y^2-x^2}{y-x}=y+x< z+y=\dfrac{z^2-y^2}{z-y}=\dfrac{f(z)-f(y)}{z-y}
$$
so where is the problem?
The problem is with the OP's conclusion that
ChiralSuperfields said:
f′≥0 for all t
It is what my post (#2) says:
Hill said:
This conclusion is incorrect.
 
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  • #5
No, You said ...
Hill said:
This conclusion is incorrect.
... so how in the world could anybody know what you meant by "this", especially if you're not sure anyway?
Hill said:
Consider the counterexample: ##f=x^2##.
... which is no counterexample. It is in fact the generic example: one has to consider the zeros of ##f''(t)## and place ##x,y,z## among possible zero(s).
 
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  • #6
fresh_42 said:
how in the world could anybody know what you meant by "this"
By looking at the quote just above it:

1717386526684.png


I give a counterexample to "this".
 
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  • #7
Hill said:
By looking at the quote just above it:

View attachment 346378

I give a counterexample to "this".
My bad (eyesight), I read it as
f''(t) ≥ 0
.
 
Last edited:
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  • #8
By the mean value theorem, there exist [itex]\zeta \in (x,y)[/itex] and [itex]\eta \in (y, z)[/itex] such that [tex]\begin{split}f'(\zeta) &= \frac{f(y) - f(x)}{y - x} \\
f'(\eta) &= \frac{f(z) - f(y)}{z - y}.
\end{split}[/tex] Now use the fact that [itex]\zeta < \eta[/itex] and [itex]f'[/itex] is increasing.
 
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  • #9
fresh_42 said:
My bad (eyesight), I read it as
.
Not entirely your fault. This is what happens when the PF "Insert quotes" feature is used to quote a LaTeX expression and the expression is not edited to be displayed in LaTeX.

You get this:
ChiralSuperfields said:
f′≥0 for all t

rather than this:
ChiralSuperfields said:
##\displaystyle f ' \geq 0## for all t

(Adding a small space, further clarifies things.)
ChiralSuperfields said:
##\displaystyle f\, ' \geq 0## for all t
 
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FAQ: Proof given ##x < y < z## and a twice differentiable function

What does it mean for a function to be twice differentiable?

A twice differentiable function is one that has both a first derivative and a second derivative that exist and are continuous over a certain interval. This implies that the function is smooth and has no abrupt changes in its slope or curvature within that interval.

How can I use the condition ##x < y < z## in analyzing a twice differentiable function?

The condition ##x < y < z## allows us to apply the Mean Value Theorem and its extensions, which can help in understanding the behavior of the function between these points. It provides a framework for discussing the function's derivatives and their relationships at these specific points.

What is the significance of the second derivative in this context?

The second derivative of a function provides information about its concavity and the nature of its critical points. If the second derivative is positive, the function is concave up at that point, indicating a local minimum. If it is negative, the function is concave down, indicating a local maximum. This information is crucial when analyzing the behavior of the function between the points ##x, y,## and ##z##.

Can I conclude anything about the function's behavior between ##x, y,## and ##z##?

Yes, by using the properties of the first and second derivatives, you can infer information about the function's increasing or decreasing behavior as well as its concavity between these points. For example, if the first derivative changes sign, it indicates a local extremum, while the second derivative can confirm the nature of that extremum.

What are some common applications of analyzing twice differentiable functions?

Common applications include optimization problems, where finding local maxima or minima is essential, and in physics, where the behavior of motion can be modeled using twice differentiable functions. They are also used in economics for analyzing cost and revenue functions to determine profit maximization points.

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