Question

If f is a differentiable function, find the value of the limit..

Answer 1

I can't tell if the the answer is either 16f'(x) or 0

Answer 2

Using a calculator, I got 16f'(x) but I wonder why it's a derivative of f there. When I tried again I did get an f'(x) in it but it's 4/h(f'(x)) I'll see what I can do again

Answer 3

Alright I got it. Here let me type it down

Answer 4

@wolfe8 the answer is 16f'(x) but I just don't understand how to get that answer manually

Answer 5

I will just focus on the function for now(will not type the limit). First rewrite it as \[\frac{ 8(f(x+h) -f(x-h)) }{ h }\]

Then, please remember that \[\frac{ f(x+h)-f(x) }{ h } =f'(x)\]

However we have (f(x+h) - f(x+h))/h which means that it is the numerator is the difference between a step and another step 2 increments above it(i.e from x-h to x+h instead of x to x+h)

Thus to get the gradient, you will have to divide it by 2h. So you will have to multiply the whole function by 2 and get \[\frac{ 16(f(x+h) - f(x+h)) }{ 2h }\]

Now you can see that it can be rewritten as 16f'(x)

After writing this I realize that maybe it doesn't make sense because it is supposed to be 16/2h * f'(x) but I think I'm close.

Answer 6

Okay now I think that I did it right because 16f'(x) also means 16 times the thing with the 2h. Ok yup I think it's right.

Answer 7

I don't know where you got the 16 and then divided everything by 2h

Answer 8

@wolfe8

Answer 9

Okay now I think that I did it right because 16f'(x) also means 16 times the thing with the 2h. Ok yup I think it's right.

Answer 10

Okay so do you know why f'(x) = \[\frac{ f(x+h)-f(x) }{h }\]? If you put values in, it is basically saying \[(y _{2}-y _{1})/(x _{2}-x _{1})\] where h = difference between 2 x points.

Answer 11

I hope this can help you understand better. Remember that gradient is difference in y divided by difference in x. I have to go now. Good luck buddy.