Is the Function Differentiable and Continuous Based on Its Derivative?

[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?

 

[what]

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

 

[dextercioby]

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.

 

[Andrew Mason]

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

 

[hypermorphism]

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.

 

[UrbanXrisis]

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?

 

[hypermorphism]

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.

 

[HallsofIvy]

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.