# Show that F'(x) exists for all x $$\in[a,b]$$

• irresistible
In summary, we need to show that F'(x) exists for all values of x in the interval [a,b], but that F':[0,1] \rightarrow1 is not integrable. To do this, we need to prove that the derivative exists at x=0, which can be done by showing that the limit \lim_{h\to 0}\frac{f(h) - f(0)}{h} exists. The derivative function, F'(x), can then be found using the product and chain rules. To prove that F'(x) is unbounded, we can take x= 1/\sqrt{2n\pi} and show that as n approaches infinity, F'(x) also
irresistible

## Homework Statement

Consider
F(x) = x2 sin(1/x2) if 0<x$$\leq1$$
and = 0 if x$$\leq0$$

Show that F'(x) exists for all x $$\in[a,b]$$ but that F':[0,1] $$\rightarrow1$$ is not integrable.

## Homework Equations

So we have to show we do not have F(1)-F(0) = $$\int$$ F'(x)dx
(integral going from 0 to 1)

## The Attempt at a Solution

I'm having trouble proving this statement.
Where should I start?
To show that F'(x) exists, should I just take the derivative or do I have to go under some long theorems of analysis to PROVE?

Most likely you'll have to actually *prove* that this derivative exists - it clearly exists for all x other than 0 because $$x^{2}sin(\frac{1}{x^{2}})$$ is the composition of differentiable functions. The only tricky part is whether F'(0) exists. To show that it does indeed exist you'll need to show that the following limit exists:

$$\lim_{h\to 0}\frac{f(h) - f(0)}{h}$$

But this just equals

$$\lim_{h\to 0}\frac{h^{2}sin(\frac{1}{h^{2}})}{h} = lim_{h\to 0}hsin(\frac{1}{h^{2}}) = 0$$

Then by application of the product rule and the chain rule,

$$F'(x) = 2xsin(\frac{1}{x^{2}}) - \frac{2cos(\frac{1}{x^{2}})}{x^{2}}$$

if x is anything other than 0, and 0 if x=0. So all you need to show is that the above function is unbounded and it will consequently not be integrable.

right,
But I think you made a little mistake with differentiation,
I think it will be:

F'(x)= 2x sin(1/x2) - 2cos(1/x2)/ x

Which still, limits exists.

But How do I prove that F'(x) is unbounded?

Take $x= 1/\sqrt{2n\pi}$. Then $1/x^2= 2n\pi$ so $sin(1/x^2)= 0$ and $cos(1/x^2)= 1$. $F'(x)= F'(1/\sqrt{2n\pi})= -2\sqrt{2n\pi}$. x will go to 0 as n goes to infinity. What happens to F'(x)?

F'(x) will be Undefined as x approaches zero $$\rightarrow$$ unbounded $$\rightarrow$$ non integrable ?

HallsofIvy just defined F'(x) above, so it *will* be defined; however, as n-->infinity, F'(x) also goes to infinity and this implies that the derivative is unbounded.

Also, thanks for pointing out my mistake; I was in an Econ class :P

## 1. What does it mean for F'(x) to exist?

For a function F(x) to have a derivative F'(x), it means that at every point x in its domain, the slope of the tangent line to the graph of F(x) can be determined. In other words, the function is smooth and continuous, with no sharp turns or breaks.

## 2. Why is it important for F'(x) to exist?

The existence of a derivative F'(x) allows us to analyze the behavior of a function and make predictions about its values. It also helps us find the maximum and minimum points of a function, which is useful in optimization problems.

## 3. How can we show that F'(x) exists for all x in [a,b]?

To show that F'(x) exists for all x in [a,b], we need to prove that the limit of the difference quotient, which represents the slope of the tangent line, exists and is the same from both the left and right sides of each point in the interval [a,b]. This can be done using the definition of the derivative or by using established theorems such as the Mean Value Theorem.

## 4. What happens if F'(x) does not exist for a certain value of x in [a,b]?

If F'(x) does not exist for a certain value of x in [a,b], it means that the function is not differentiable at that point. This could happen if the function has a sharp corner or a discontinuity at that point. In this case, we cannot determine the slope of the tangent line and the function cannot be analyzed using calculus methods.

## 5. Can we use the existence of F'(x) to determine the continuity of a function?

No, the existence of F'(x) does not guarantee the continuity of a function. A function can be differentiable at a point but still have a discontinuity or a sharp corner at that point, which would make it not continuous. However, if a function is continuous at a point, then it is also differentiable at that point.

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