# Difinition of a derivative

• UrbanXrisis
In summary: This is a piecewise function. f(x) has one rule when x is not zero (!=0), and another rule when x=0.wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x can't equal zero?No, he didn't say that. He said that as long as x is NOT 0, you define f(x) to be x2 sin(1/x) (which obviously can't be correct for x=0) but that if x IS 0, f(0)= 0.
UrbanXrisis
let f be a function such that $$lim_{ h->x} \frac{f(2+h)-f(2)}{h} = 5$$

Which of the following must be true?

I. f is continuous at x=2.
II. F is differentiable at x=2.
III. The derivative of f is continuous at x=2.

I know (I) is true because it can be differentiated
I know that (II) is ture to because the derivative was found so it can be differentiable.
I don't know if III is true because the it doesn't tell me the limit of f'(x)

are these correct?

That isn't the limit definition for a derivative unless x = 0, or maybe you typed it in wrong.

UrbanXrisis said:
let f be a function such that $$lim_{ h->0} \frac{f(2+h)-f(2)}{h} = 5$$

Which of the following must be true?

I. f is continuous at x=2.
II. F is differentiable at x=2.
III. The derivative of f is continuous at x=2.

I know (I) is true because it can be differentiated.I know that (II) is ture to because the derivative was found so it can be differentiable.
I don't know if (III) is true because the it doesn't tell me the limit of f'(x)

are these correct?

It's okay...At least,for me...I edited your typo and advise u to use the code

\lim_{...} for the limit and \rightarrow for the "->"...

Daniel.

UrbanXrisis said:
let f be a function such that $$lim_{ h->x} \frac{f(2+h)-f(2)}{h} = 5$$

Which of the following must be true?

I. f is continuous at x=2.
II. F is differentiable at x=2.
III. The derivative of f is continuous at x=2.
I think this must be:

$$\lim_{h\rightarrow{0}} \frac{f(2+h)-f(2)}{h} = 5$$ (ie. 0, not x)

I know (I) is true because it can be differentiated
I agree, with one caveat: We don't know the domain of h. If h can be positive or negative, f would be differentiable at 2 and, therefore, continous.

I know that (II) is ture to because the derivative was found so it can be differentiable.
True, subject to the above.
I don't know if III is true because the it doesn't tell me the limit of f'(x)
I think that if a function f(x) is differentiable at x, its derivative must be continuous at x.

AM

Andrew Mason said:
I think that if a function f(x) is differentiable at x, its derivative must be continuous at x.
AM

Not true. Counterexample: Let f(x) = x2sin(1/x) for x != 0, and f(x) = 0 when x = 0. Then f is differentiable at zero, but the derivative is not continuous at 0.

hypermorphism said:
Not true. Counterexample: Let f(x) = x2sin(1/x) for x != 0, and f(x) = 0 when x = 0. Then f is differentiable at zero, but the derivative is not continuous at 0.

wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x can't equal zero?

UrbanXrisis said:
wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x can't equal zero?

This is a piecewise function. f(x) has one rule when x is not zero (!=0), and another rule when x=0.

UrbanXrisis said:
wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x can't equal zero?

No, he didn't say that. He said that as long as x is NOT 0, you define
f(x) to be x2 sin(1/x) (which obviously can't be correct for x=0) but that if x IS 0, f(0)= 0.

## What is the definition of a derivative?

The derivative of a function at a given point is the slope of the tangent line to the graph of the function at that point.

## How is the derivative of a function calculated?

The derivative of a function can be calculated using the formula: f'(x) = lim(h->0) (f(x+h)-f(x))/h, where h represents the change in x.

## What does the derivative represent?

The derivative represents the rate of change of a function at a specific point. It can also be interpreted as the instantaneous rate of change or the slope of the function at that point.

## Why is the derivative important?

The derivative is important because it allows us to analyze the behavior of a function and understand how it changes at a particular point. It is also used to solve optimization problems and to find the maximum and minimum values of a function.

## Can the derivative be negative?

Yes, the derivative can be negative. This means that the function is decreasing at that point. A positive derivative indicates that the function is increasing at that point, while a derivative of zero means that the function is neither increasing nor decreasing at that point.

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