I'm confused with this question: If f(a)=0 and f'(a)= 4, find lim h->0 f(a+h)/2h

Answer 1

f'(a) = 4

f(a) = 4x (or 4x + c, but the constant drops out so it doesn't really matter if you treat c as zero or some other constant)

f(a+h) = 4(x+h) = 4x+4h

[f(a+h) - f(a)]/h = [4x+4h -4x]/h = 4h/h = 4

lim as h goes to zero of [f(a+h) - f(a)]/h = 4 which is the same as f'(a), the derivative of f(a)

Answer 2

Hi Sophie, yes it's confusing because there are few things you have to consider. Let's have first the definition of derivative in terms of limits:

f'(x) = lim_h→0 (f(x+h)-f(x))/h

Let's replace x with a. So we'll have:

f'(a) = lim_h→0 (f(a+h)-f(a))/h

Now, let's consider some of the given:

f(a) = 0

f'(a) = 4

and we are looking for lim_h→0 f(a+h)/2h

Substitute f(a) and f'(a) from the formula:

4 = lim_h→0 (f(a+h)-0)/h

4 = lim_h→0 f(a+h)/h

or lim_h→0 f(a+h)/h = 4

multiply 1/2 on both sides:

(1/2)lim_h→0 f(a+h)/h = 2

which is the same as:

lim_h→0 f(a+h)/2h = 2